Q. No.Find the magnitude of lattice energy of LiF (in kJ/mol):
Given :
Enthalpy of sublimation of Li(s) = 161 kJ/mol
Ionisation enthalpy of Li(g) = 520 kJ/mol
Bond Enthalpy of \(F_2\) (g) = 154 kJ/mol
Electron gain enthalpy of F(g) = –328 kJ/mol
Enthalpy of formation of LiF(s) = – 617 kJ/mol
1. -947
2. -1047
3. +956
4. +1097
Given :
Enthalpy of sublimation of Li(s) = 161 kJ/mol
Ionisation enthalpy of Li(g) = 520 kJ/mol
Bond Enthalpy of \(F_2\) (g) = 154 kJ/mol
Electron gain enthalpy of F(g) = –328 kJ/mol
Enthalpy of formation of LiF(s) = – 617 kJ/mol
1. -947
2. -1047
3. +956
4. +1097
Subtopic: Hess's Law | Thermochemistry |
68%
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The bond dissociation enthalpy of X2 \(\Delta \mathrm{H}_{\mathrm{bond}}^{\mathrm{o}}\) in kJ mol–1 calculated from the given data is:
\(\begin{aligned} &\small \mathrm{MX}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) ,\Delta \mathrm{H}_{\text {lattice }}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}), \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})^{-}+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\small \mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) ,\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)
[Given: M+X – is a pure ionic compound and X forms a diatomic
molecule X2 in the gaseous state]
1. 100
2. 150
3. 200
4. 250
\(\begin{aligned} &\small \mathrm{MX}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) ,\Delta \mathrm{H}_{\text {lattice }}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}), \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})^{-}+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\small \mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) ,\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)
[Given: M+X – is a pure ionic compound and X forms a diatomic
molecule X2 in the gaseous state]
1. 100
2. 150
3. 200
4. 250
Subtopic: Hess's Law | Thermochemistry |
65%
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Consider the following reactions:
\(\mathrm{S}_{(\mathrm{~g})}+\frac{3}{2} \mathrm{{O}_2}_{(\mathrm{~g})} \rightarrow \mathrm{{SO}_3}_{(\mathrm{~g})}+2 \mathrm{x} \mathrm{~kcal}\)
\(\mathrm{{SO}_2}_{(\mathrm{~g})}+\frac{1}{2} \mathrm{{O}_2}_{(\mathrm{~g})} \rightarrow \mathrm{{SO}_3}_{(\mathrm{~g})}+\mathrm{y} \mathrm{~kcal}\)
The heat of formation of SO2(g) is given by:
\(\mathrm{S}_{(\mathrm{~g})}+\frac{3}{2} \mathrm{{O}_2}_{(\mathrm{~g})} \rightarrow \mathrm{{SO}_3}_{(\mathrm{~g})}+2 \mathrm{x} \mathrm{~kcal}\)
\(\mathrm{{SO}_2}_{(\mathrm{~g})}+\frac{1}{2} \mathrm{{O}_2}_{(\mathrm{~g})} \rightarrow \mathrm{{SO}_3}_{(\mathrm{~g})}+\mathrm{y} \mathrm{~kcal}\)
The heat of formation of SO2(g) is given by:
| 1. | \(\dfrac{2 \mathrm{x}}{\mathrm{y}} \mathrm{kcal}\) | 2. | y - 2x kcal |
| 3. | 2x + y kcal | 4. | x + y kcal |
Subtopic: Thermochemistry |
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From the given data, find the enthalpy of hydrogenation of ethene in kJ/mol.
(a) B.E. of C – C = 350 kJ/mol
(b) B.E. of C = C = 600 kJ/mol
(c) B.E. of H – H = 400 kJ/mol
(d) B.E. of C – H = 410 kJ/mol
1. –170
2. –580
3. +170
4. +580
(a) B.E. of C – C = 350 kJ/mol
(b) B.E. of C = C = 600 kJ/mol
(c) B.E. of H – H = 400 kJ/mol
(d) B.E. of C – H = 410 kJ/mol
1. –170
2. –580
3. +170
4. +580
Subtopic: Thermochemistry |
78%
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Calculate the standard enthalpy of formation (\(\Delta_fH^\circ\)) for 2 moles of liquid benzene (\(\mathrm{{C_6H_6}_{(l)}}\)) at 25°C, based on the given thermodynamic data.
Given Data:
\(\Delta _c\mathrm H\mathrm{(C_6H_6}_\mathrm{(l)})~= -3264.6~ \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta _c\mathrm H\mathrm{(C}_\mathrm{(s)})~= -393.5~ \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta _f\mathrm H\mathrm{(H_2O}_\mathrm{(l)})~= -285.83~ \mathrm{~kJ} / \mathrm{mol}\)
1. \(-~92.22 \mathrm{~kJ} / \mathrm{mol}\)
2. \(-46.11 \mathrm{~kJ} / \mathrm{mol}\)
3. \(+~92.22 \mathrm{~kJ} / \mathrm{mol}\)
4. \(+46.11 \mathrm{~kJ} / \mathrm{mol}\)
Given Data:
\(\Delta _c\mathrm H\mathrm{(C_6H_6}_\mathrm{(l)})~= -3264.6~ \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta _c\mathrm H\mathrm{(C}_\mathrm{(s)})~= -393.5~ \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta _f\mathrm H\mathrm{(H_2O}_\mathrm{(l)})~= -285.83~ \mathrm{~kJ} / \mathrm{mol}\)
1. \(-~92.22 \mathrm{~kJ} / \mathrm{mol}\)
2. \(-46.11 \mathrm{~kJ} / \mathrm{mol}\)
3. \(+~92.22 \mathrm{~kJ} / \mathrm{mol}\)
4. \(+46.11 \mathrm{~kJ} / \mathrm{mol}\)
Subtopic: Thermochemistry |
Level 3: 35%-60%
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What is the value of enthalpy change \( (\Delta\text{H})\) (in kJ/mole) for the given reaction?
\(3 \mathrm{C}(\mathrm{s})+\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})\)
Given :
\(\begin{aligned} & 2 \mathrm{Fe}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s}) \Delta \mathrm{H}^{\circ}=-824 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\)
1. 494
2. 984
3. 700
4. 560
\(3 \mathrm{C}(\mathrm{s})+\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})\)
Given :
\(\begin{aligned} & 2 \mathrm{Fe}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s}) \Delta \mathrm{H}^{\circ}=-824 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\)
1. 494
2. 984
3. 700
4. 560
Subtopic: Thermochemistry |
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Hints
Equal volume of \(1~ \mathrm{M}~ \mathrm{HCl}\) and \(1~ \mathrm{M}~ \mathrm{H}_2 \mathrm{SO}_4\) neutralised by dil. \(\mathrm{NaOH}\) and heat released is \(\mathrm{x}\) and \(\mathrm{y}\) KCal \(mol^{-1}\) respectively. Which of the following is the correct interpretation?
1. \( x=y \)
2. \(x=0.5 y \)
3. \(x=0.4 y \)
4. \( x=2 y\)
1. \( x=y \)
2. \(x=0.5 y \)
3. \(x=0.4 y \)
4. \( x=2 y\)
Subtopic: Thermochemistry |
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Heat of solution of \(\mathrm{CuSO}_4\) and \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\) is given as (–70 kJ/mole) and (+12 kJ/mole) respectively. Then \(\Delta \mathrm{H}_{\text {hydration }}\) for converting \(\mathrm{CuSO}_4\) to \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\) is (–x kJ/mole). Find the value of x.
1. 82
2. 98
2. 67
4. 90
1. 82
2. 98
2. 67
4. 90
Subtopic: Thermochemistry |
85%
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If the standard enthalpy of vaporization of \( \text{CCl}_4\) is 30.5 kJ/mol, find heat absorbed for vaporization of 294 gm of \( \text{CCl}_4.\)
1. 78 KJ
2. 58 KJ
3. 100 KJ
4. 80 KJ
1. 78 KJ
2. 58 KJ
3. 100 KJ
4. 80 KJ
Subtopic: Thermochemistry |
85%
Level 1: 80%+
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An athlete is given 100 g of glucose energy equivalent to 1560 kJ to utilize 50 % of this gained energy in an event. Enthalpy of evaporation of H2O is 44 kJ/mol. In order to avoid storage of energy in the body the mass of water (in g) he would perspire is:
(Round off the nearest Integer)
1. 325 g
2. 319 g
3. 298 g
4. 345 g
(Round off the nearest Integer)
1. 325 g
2. 319 g
3. 298 g
4. 345 g
Subtopic: Thermochemistry |
67%
Level 2: 60%+
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