Classes

First Year Chemistry ThermoChemistry Q.20 What is the meaning of the term enthalpy of ionization? If the heat of neutralization of \mathrm{HCl} and \mathrm{NaOH} is -57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} and heat of neutralization o


Change the way you learn with Maqsad's classes. Local examples, engaging animations, and instant video solutions keep you on your toes and make learning fun like never before!

Class 9Class 10First YearSecond Year
Q.20 What is the meaning of the term enthalpy of ionization? If the heat of neutralization of \mathrm{HCl} and \mathrm{NaOH} is -57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} and heat of neutralization of \mathrm{CH}_{3} \mathrm{COOH} with \mathrm{NaOH} is -55.2 \mathrm{~kJ} \mathrm{~mol}^{-1} calculate the enthalpy of ionization of \mathrm{CH}_{3} \mathrm{COOH} .(Ans: 2.1 \mathrm{~kJ} \mathrm{~mol}^{-1} )

Q.5(a) Differentiate between the following:(i) Internal energy and enthalpy
Q.5(a) Differentiate between the following:(i) Internal energy and enthalpy

Q.5(a) Differentiate between the following:(i) Internal energy and enthalpy

Q.4. Define the following terms and give three examples of each(v) Exothermic reaction
Q.4. Define the following terms and give three examples of each(v) Exothermic reaction

Q.4. Define the following terms and give three examples of each(v) Exothermic reaction

Q.4. Define the following terms and give three examples of each(viii) Enthalpy of the system
Q.4. Define the following terms and give three examples of each(viii) Enthalpy of the system

Q.4. Define the following terms and give three examples of each(viii) Enthalpy of the system

(b) How do you measure the heat of combustion of a substance by bomb calorimeter.
(b) How do you measure the heat of combustion of a substance by bomb calorimeter.

(b) How do you measure the heat of combustion of a substance by bomb calorimeter.

Q.4. Define the following terms and give three examples of each(ii) Surroundings
Q.4. Define the following terms and give three examples of each(ii) Surroundings

Q.4. Define the following terms and give three examples of each(ii) Surroundings

Q.20 What is the meaning of the term enthalpy of ionization? If the heat of neutralization of \mathrm{HCl} and \mathrm{NaOH} is -57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} and heat of neutralization of \mathrm{CH}_{3} \mathrm{COOH} with \mathrm{NaOH} is -55.2 \mathrm{~kJ} \mathrm{~mol}^{-1} calculate the enthalpy of ionization of \mathrm{CH}_{3} \mathrm{COOH} .(Ans: 2.1 \mathrm{~kJ} \mathrm{~mol}^{-1} )
Q.20 What is the meaning of the term enthalpy of ionization? If the heat of neutralization of  \mathrm{HCl}  and  \mathrm{NaOH}  is  -57.3 \mathrm{~kJ} \mathrm{~mol}^{-1}  and heat of neutralization of  \mathrm{CH}_{3} \mathrm{COOH}  with  \mathrm{NaOH}  is  -55.2 \mathrm{~kJ} \mathrm{~mol}^{-1}  calculate the enthalpy of ionization of  \mathrm{CH}_{3} \mathrm{COOH} .(Ans:  2.1 \mathrm{~kJ} \mathrm{~mol}^{-1}  )
now playing

Q.20 What is the meaning of the term enthalpy of ionization? If the heat of neutralization of \mathrm{HCl} and \mathrm{NaOH} is -57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} and heat of neutralization of \mathrm{CH}_{3} \mathrm{COOH} with \mathrm{NaOH} is -55.2 \mathrm{~kJ} \mathrm{~mol}^{-1} calculate the enthalpy of ionization of \mathrm{CH}_{3} \mathrm{COOH} .(Ans: 2.1 \mathrm{~kJ} \mathrm{~mol}^{-1} )

(b) Define the following enthalpies and give two examples of each.(iv)Standard enthalpy of solution
(b) Define the following enthalpies and give two examples of each.(iv)Standard enthalpy of solution

(b) Define the following enthalpies and give two examples of each.(iv)Standard enthalpy of solution

Q.16 By applying Hesss law calculate the enthalpy change for the formation of an. aqueous solution of \mathrm{NH}_{4} \mathrm{Cl} from \mathrm{NH}_{3} gas and \mathrm{HCl} gas. The results for the various reactions are as follows:(i) \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{aq} \rightarrow \mathrm{NH}_{3}(\mathrm{aq}) \Delta \mathrm{H}=-35.16 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii) \mathrm{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow \mathrm{HCl} (aq) \Delta \mathrm{H}=-72.41 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii) \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{HCl} (aq) \rightarrow \mathrm{NH}_{4} \mathrm{Cl} (aq) \Delta \mathrm{H}=-51.48 \mathrm{~kJ} \mathrm{~mol}^{-1}
Q.16 By applying Hesss law calculate the enthalpy change for the formation of an. aqueous solution of  \mathrm{NH}_{4} \mathrm{Cl}  from  \mathrm{NH}_{3}  gas and  \mathrm{HCl}  gas. The results for the various reactions are as follows:(i)   \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{aq} \rightarrow \mathrm{NH}_{3}(\mathrm{aq})  \Delta \mathrm{H}=-35.16 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii)  \mathrm{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow \mathrm{HCl}  (aq) \Delta \mathrm{H}=-72.41 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii)   \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{HCl}  (aq)  \rightarrow \mathrm{NH}_{4} \mathrm{Cl}  (aq) \Delta \mathrm{H}=-51.48 \mathrm{~kJ} \mathrm{~mol}^{-1}

Q.16 By applying Hesss law calculate the enthalpy change for the formation of an. aqueous solution of \mathrm{NH}_{4} \mathrm{Cl} from \mathrm{NH}_{3} gas and \mathrm{HCl} gas. The results for the various reactions are as follows:(i) \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{aq} \rightarrow \mathrm{NH}_{3}(\mathrm{aq}) \Delta \mathrm{H}=-35.16 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii) \mathrm{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow \mathrm{HCl} (aq) \Delta \mathrm{H}=-72.41 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii) \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{HCl} (aq) \rightarrow \mathrm{NH}_{4} \mathrm{Cl} (aq) \Delta \mathrm{H}=-51.48 \mathrm{~kJ} \mathrm{~mol}^{-1}

Q.1 Select the suitable answer from the given choices.(viii) The net heat change in a chemical reaction is same whether it is brought about in two or more different ways in one or several steps. It is known as(a) Henrys law.(b) Joules principle(c) Hesss law(d) Law of conservation of energy
Q.1 Select the suitable answer from the given choices.(viii) The net heat change in a chemical reaction is same whether it is brought about in two or more different ways in one or several steps. It is known as(a) Henrys law.(b) Joules principle(c) Hesss law(d) Law of conservation of energy

Q.1 Select the suitable answer from the given choices.(viii) The net heat change in a chemical reaction is same whether it is brought about in two or more different ways in one or several steps. It is known as(a) Henrys law.(b) Joules principle(c) Hesss law(d) Law of conservation of energy

(b) Hesss law helps us to calculate the heats of those reactions which cannot be normally carried out in a laboratory. Explain it.
(b) Hesss law helps us to calculate the heats of those reactions which cannot be normally carried out in a laboratory. Explain it.

(b) Hesss law helps us to calculate the heats of those reactions which cannot be normally carried out in a laboratory. Explain it.

(b) Define the following enthalpies and give two examples of each.(ii) Standard enthalpy of combustion
(b) Define the following enthalpies and give two examples of each.(ii) Standard enthalpy of combustion

(b) Define the following enthalpies and give two examples of each.(ii) Standard enthalpy of combustion

(c) Using the information given in the table below calculate the lattice energy of potassium bromide.\begin{tabular}{|c|c|}\hline Reactions & \Delta \mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1} \\\hline \mathrm{K}(\mathrm{s})+1 / 2 \mathrm{Br}_{2}() \rightarrow \mathrm{K}^{+} \mathrm{Br}^{-}(\mathrm{s}) & -392 \\ \mathrm{~K}(\mathrm{~s}) \rightarrow \mathrm{K}(\mathrm{g}) & +90 \\ \mathrm{~K}(\mathrm{~g}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{e}^{-} & +420 \\ 1 / 2 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Br}(\mathrm{g}) & +112 \\ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g}) & -342 \\\hline\end{tabular}
(c) Using the information given in the table below calculate the lattice energy of potassium bromide.\begin{tabular}{|c|c|}\hline Reactions &  \Delta \mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1}  \\\hline  \mathrm{K}(\mathrm{s})+1 / 2 \mathrm{Br}_{2}() \rightarrow \mathrm{K}^{+} \mathrm{Br}^{-}(\mathrm{s})  &  -392  \\ \mathrm{~K}(\mathrm{~s}) \rightarrow \mathrm{K}(\mathrm{g})  &  +90  \\ \mathrm{~K}(\mathrm{~g}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{e}^{-}  &  +420  \\ 1 / 2 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Br}(\mathrm{g})  &  +112  \\ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g})  &  -342  \\\hline\end{tabular}

(c) Using the information given in the table below calculate the lattice energy of potassium bromide.\begin{tabular}{|c|c|}\hline Reactions & \Delta \mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1} \\\hline \mathrm{K}(\mathrm{s})+1 / 2 \mathrm{Br}_{2}() \rightarrow \mathrm{K}^{+} \mathrm{Br}^{-}(\mathrm{s}) & -392 \\ \mathrm{~K}(\mathrm{~s}) \rightarrow \mathrm{K}(\mathrm{g}) & +90 \\ \mathrm{~K}(\mathrm{~g}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{e}^{-} & +420 \\ 1 / 2 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Br}(\mathrm{g}) & +112 \\ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g}) & -342 \\\hline\end{tabular}

Q.4. Define the following terms and give three examples of each(iii) State function
Q.4. Define the following terms and give three examples of each(iii) State function

Q.4. Define the following terms and give three examples of each(iii) State function

(c) Using the information given in the table below calculate the lattice energy of potassium bromide.\begin{tabular}{|c|c|}\hline Reactions & \Delta \mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1} \\\hline \mathrm{K}(\mathrm{s})+1 / 2 \mathrm{Br}_{2}() \rightarrow \mathrm{K}^{+} \mathrm{Br}^{-}(\mathrm{s}) & -392 \\ \mathrm{~K}(\mathrm{~s}) \rightarrow \mathrm{K}(\mathrm{g}) & +90 \\ \mathrm{~K}(\mathrm{~g}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{e}^{-} & +420 \\ 1 / 2 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Br}(\mathrm{g}) & +112 \\ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g}) & -342 \\\hline\end{tabular}
(c) Using the information given in the table below calculate the lattice energy of potassium bromide.\begin{tabular}{|c|c|}\hline Reactions &  \Delta \mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1}  \\\hline  \mathrm{K}(\mathrm{s})+1 / 2 \mathrm{Br}_{2}() \rightarrow \mathrm{K}^{+} \mathrm{Br}^{-}(\mathrm{s})  &  -392  \\ \mathrm{~K}(\mathrm{~s}) \rightarrow \mathrm{K}(\mathrm{g})  &  +90  \\ \mathrm{~K}(\mathrm{~g}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{e}^{-}  &  +420  \\ 1 / 2 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Br}(\mathrm{g})  &  +112  \\ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g})  &  -342  \\\hline\end{tabular}

(c) Using the information given in the table below calculate the lattice energy of potassium bromide.\begin{tabular}{|c|c|}\hline Reactions & \Delta \mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1} \\\hline \mathrm{K}(\mathrm{s})+1 / 2 \mathrm{Br}_{2}() \rightarrow \mathrm{K}^{+} \mathrm{Br}^{-}(\mathrm{s}) & -392 \\ \mathrm{~K}(\mathrm{~s}) \rightarrow \mathrm{K}(\mathrm{g}) & +90 \\ \mathrm{~K}(\mathrm{~g}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{e}^{-} & +420 \\ 1 / 2 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Br}(\mathrm{g}) & +112 \\ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(\mathrm{g}) & -342 \\\hline\end{tabular}

Q.6(a) What are spontaneous and non-spontaneous processes. Give examples.
Q.6(a) What are spontaneous and non-spontaneous processes. Give examples.

Q.6(a) What are spontaneous and non-spontaneous processes. Give examples.

Q.19 Graphite and diamond are two forms of carbon. The enthalpy of combustion of graphite at 25^{\circ} \mathrm{C} is -393.51 \mathrm{~kJ} mol-1 and that of diamond is -395.41 \mathrm{~kJ} \mathrm{~mol}^{-1} . What is the enthalpy change of the process? Graphite \rightarrow Diamond at the same temperature? \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) (Ans: 1.91 \mathrm{~kJ} \mathrm{~mol}^{-1} )
Q.19 Graphite and diamond are two forms of carbon. The enthalpy of combustion of graphite at  25^{\circ} \mathrm{C}  is  -393.51 \mathrm{~kJ}  mol-1 and that of diamond is  -395.41 \mathrm{~kJ} \mathrm{~mol}^{-1} . What is the enthalpy change of the process? Graphite  \rightarrow  Diamond at the same temperature? \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) (Ans:  1.91 \mathrm{~kJ} \mathrm{~mol}^{-1}  )

Q.19 Graphite and diamond are two forms of carbon. The enthalpy of combustion of graphite at 25^{\circ} \mathrm{C} is -393.51 \mathrm{~kJ} mol-1 and that of diamond is -395.41 \mathrm{~kJ} \mathrm{~mol}^{-1} . What is the enthalpy change of the process? Graphite \rightarrow Diamond at the same temperature? \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) (Ans: 1.91 \mathrm{~kJ} \mathrm{~mol}^{-1} )

Q.1 Select the suitable answer from the given choices.(ii) In endothermic reactions the heat content of the:(a) products is more than that of reactants(b) reactants is more than that of products(c) surroundings in creases(d) reactants and products is equal
Q.1 Select the suitable answer from the given choices.(ii) In endothermic reactions the heat content of the:(a) products is more than that of reactants(b) reactants is more than that of products(c) surroundings in creases(d) reactants and products is equal

Q.1 Select the suitable answer from the given choices.(ii) In endothermic reactions the heat content of the:(a) products is more than that of reactants(b) reactants is more than that of products(c) surroundings in creases(d) reactants and products is equal

Q.1 Select the suitable answer from the given choices.(iv) The change in heat energy of a chemical reaction at constant temperature and pressure is called:(a) enthalpy change(c) heat of sublimation(b) bondenergy(d) internal energy change
Q.1 Select the suitable answer from the given choices.(iv) The change in heat energy of a chemical reaction at constant temperature and pressure is called:(a) enthalpy change(c) heat of sublimation(b) bondenergy(d) internal energy change

Q.1 Select the suitable answer from the given choices.(iv) The change in heat energy of a chemical reaction at constant temperature and pressure is called:(a) enthalpy change(c) heat of sublimation(b) bondenergy(d) internal energy change

(b)Explain that burning of a candle is a spontaneous process.
(b)Explain that burning of a candle is a spontaneous process.
video locked

(b)Explain that burning of a candle is a spontaneous process.

Q.4. Define the following terms and give three examples of each(iii) State function
Q.4. Define the following terms and give three examples of each(iii) State function
video locked

Q.4. Define the following terms and give three examples of each(iii) State function

Q.17 Calculate the heat of formation of ethyl alcohol from the following information(i) Heat of combustion of ethyl alcohol is -1367 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii) Heat of formation of carbon dioxide is- 393.7 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii) Heat of formation of water is -285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} (Ans:-278.4 kcal mol { }^{-1} )
Q.17 Calculate the heat of formation of ethyl alcohol from the following information(i) Heat of combustion of ethyl alcohol is  -1367 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii) Heat of formation of carbon dioxide is-  393.7 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii) Heat of formation of water is  -285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} (Ans:-278.4 kcal mol  { }^{-1}  )
video locked

Q.17 Calculate the heat of formation of ethyl alcohol from the following information(i) Heat of combustion of ethyl alcohol is -1367 \mathrm{~kJ} \mathrm{~mol}^{-1} (ii) Heat of formation of carbon dioxide is- 393.7 \mathrm{~kJ} \mathrm{~mol}^{-1} (iii) Heat of formation of water is -285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} (Ans:-278.4 kcal mol { }^{-1} )

(b) Hesss law helps us to calculate the heats of those reactions which cannot be normally carried out in a laboratory. Explain it.
(b) Hesss law helps us to calculate the heats of those reactions which cannot be normally carried out in a laboratory. Explain it.
video locked

(b) Hesss law helps us to calculate the heats of those reactions which cannot be normally carried out in a laboratory. Explain it.

Q.2 Fill in the blanks with suitable words.(i) The substance undergoing a physical or a chemical change forms a chemical.(ii) The change in internal energy be measured.(iii) Solids which have more than one crystalline forms possess values of heats of formation.(iv) A process is called if it takes place on its own without any external assistance.(v) A is a macroscopic property of a system which is of the path adopted to bring about that change.
Q.2 Fill in the blanks with suitable words.(i) The substance undergoing a physical or a chemical change forms a chemical.(ii) The change in internal energy be measured.(iii) Solids which have more than one crystalline forms possess values of heats of formation.(iv) A process is called if it takes place on its own without any external assistance.(v) A is a macroscopic property of a system which is of the path adopted to bring about that change.
video locked

Q.2 Fill in the blanks with suitable words.(i) The substance undergoing a physical or a chemical change forms a chemical.(ii) The change in internal energy be measured.(iii) Solids which have more than one crystalline forms possess values of heats of formation.(iv) A process is called if it takes place on its own without any external assistance.(v) A is a macroscopic property of a system which is of the path adopted to bring about that change.

Example 3: 10.16 \mathrm{~g} of graphite is burnt in a bomb calorimeter and the temperature rise recorded is 3.87 \mathrm{~K} . Calculate the enthalpy of combustion of graphite if the heat capacity of the calorimeter (bomb water etc.) is 86.02 \mathrm{~kJ} \mathrm{~K}^{-1}
Example 3: 10.16 \mathrm{~g}  of graphite is burnt in a bomb calorimeter and the temperature rise recorded is  3.87 \mathrm{~K} . Calculate the enthalpy of combustion of graphite if the heat capacity of the calorimeter (bomb water etc.) is  86.02 \mathrm{~kJ} \mathrm{~K}^{-1}
video locked

Example 3: 10.16 \mathrm{~g} of graphite is burnt in a bomb calorimeter and the temperature rise recorded is 3.87 \mathrm{~K} . Calculate the enthalpy of combustion of graphite if the heat capacity of the calorimeter (bomb water etc.) is 86.02 \mathrm{~kJ} \mathrm{~K}^{-1}

Q.13 50 \mathrm{~cm}^{3} of 1.0 \mathrm{MHCl} is mixed with 50 \mathrm{~cm}^{3} of 1.00 \mathrm{M} \mathrm{NaOH} in a glass calorimeter. The temperature of the resultant mixture increases from 21.0^{\circ} \mathrm{C} to 27.5^{\circ} \mathrm{C} . Assume that calorimeter losses of heat are negligible. Calculate the enthalpy change mole-1 for the reactions. The density of solution to be considered is 1 \mathrm{gcm}^{-3} and specific heat is 4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1} .(Ans: - 54 \mathrm{~kJ} \mathrm{~mol}^{-1} )
Q.13  50 \mathrm{~cm}^{3}  of  1.0 \mathrm{MHCl}  is mixed with  50 \mathrm{~cm}^{3}  of  1.00 \mathrm{M} \mathrm{NaOH}  in a glass calorimeter. The temperature of the resultant mixture increases from  21.0^{\circ} \mathrm{C}  to  27.5^{\circ} \mathrm{C} . Assume that calorimeter losses of heat are negligible. Calculate the enthalpy change mole-1 for the reactions. The density of solution to be considered is  1 \mathrm{gcm}^{-3}  and specific heat is  4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1} .(Ans: -  54 \mathrm{~kJ} \mathrm{~mol}^{-1}  )
video locked

Q.13 50 \mathrm{~cm}^{3} of 1.0 \mathrm{MHCl} is mixed with 50 \mathrm{~cm}^{3} of 1.00 \mathrm{M} \mathrm{NaOH} in a glass calorimeter. The temperature of the resultant mixture increases from 21.0^{\circ} \mathrm{C} to 27.5^{\circ} \mathrm{C} . Assume that calorimeter losses of heat are negligible. Calculate the enthalpy change mole-1 for the reactions. The density of solution to be considered is 1 \mathrm{gcm}^{-3} and specific heat is 4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1} .(Ans: - 54 \mathrm{~kJ} \mathrm{~mol}^{-1} )

(b) Draw a complete fully labeled Born-Haber cycle for the formation of potassium bromide.
(b) Draw a complete fully labeled Born-Haber cycle for the formation of potassium bromide.
video locked

(b) Draw a complete fully labeled Born-Haber cycle for the formation of potassium bromide.

(b) Define the following enthalpies and give two examples of each.(i) Standard enthalpy of reaction
(b) Define the following enthalpies and give two examples of each.(i) Standard enthalpy of reaction
video locked

(b) Define the following enthalpies and give two examples of each.(i) Standard enthalpy of reaction

When 2.00 \mathrm{~mol} of \mathrm{H}_{2} and 1.00 \mathrm{~mol} \mathrm{of} \mathrm{} \mathrm{O}_{2} at 100^{\circ} \mathrm{C} and 1 torr pressure react to produce 2.00 \mathrm{~mol} of gaseous water 484.5 \mathrm{~kJ} of energy are evolved. What are the values of (a) \Delta \mathrm{H} (b) \Delta \mathrm{E} for the production of one mole of \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ?
When  2.00 \mathrm{~mol}  of  \mathrm{H}_{2}  and  1.00 \mathrm{~mol} \mathrm{of} \mathrm{} \mathrm{O}_{2}  at  100^{\circ} \mathrm{C}  and 1 torr pressure react to produce  2.00 \mathrm{~mol}  of gaseous water  484.5 \mathrm{~kJ}  of energy are evolved. What are the values of (a)  \Delta \mathrm{H}  (b)  \Delta \mathrm{E}  for the production of one mole of  \mathrm{H}_{2} \mathrm{O}(\mathrm{g})  ?
video locked

When 2.00 \mathrm{~mol} of \mathrm{H}_{2} and 1.00 \mathrm{~mol} \mathrm{of} \mathrm{} \mathrm{O}_{2} at 100^{\circ} \mathrm{C} and 1 torr pressure react to produce 2.00 \mathrm{~mol} of gaseous water 484.5 \mathrm{~kJ} of energy are evolved. What are the values of (a) \Delta \mathrm{H} (b) \Delta \mathrm{E} for the production of one mole of \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ?

(c) Is it true that a non-spontaneous process never happens in the universe?Explain it.
(c) Is it true that a non-spontaneous process never happens in the universe?Explain it.
video locked

(c) Is it true that a non-spontaneous process never happens in the universe?Explain it.

MDCAT/ ECAT question bank