Q. No.Calculate Bond Energy of C-H bond in \(CH_4\):
Given: \(\Delta H_r\) of \(CH_4(g) = -\text {x kJ/mol}\)
\(\Delta H_{sub}\) of carbon = y kJ/mol
B.E. of H-H=z kJ/mol
1. \(\frac{x-y+z}{4} \)
2. \(\frac{y+2 z+x}{4} \)
3. \(\frac{y-2 z-x}{4} \)
4. \(\frac{2 y-z+x}{4}\)
Given: \(\Delta H_r\) of \(CH_4(g) = -\text {x kJ/mol}\)
\(\Delta H_{sub}\) of carbon = y kJ/mol
B.E. of H-H=z kJ/mol
1. \(\frac{x-y+z}{4} \)
2. \(\frac{y+2 z+x}{4} \)
3. \(\frac{y-2 z-x}{4} \)
4. \(\frac{2 y-z+x}{4}\)
Subtopic: Hess's Law |
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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 |
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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 |
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Calculate the enthalpy of formation of ethane at \(25^\circ\text{C}\) and 1 atm pressure, if the enthalpies of combustion are as follows :
1. \(+54.0~\text{kJ mol}^{–1} \)
2. \(-68.0~\text{kJ mol}^{–1} \)
3. \(-86.0~\text{kJ mol}^{–1} \)
4. \(+97.0~\text{kJ mol}^{–1} \)
| Substance | \(H_2\) | C(graphite) | \(C_2H_6(g)\) |
| \(\dfrac{\Delta_{{c}} {H}^{\ominus}}{\mathrm{kJmol}^{-1}}\) | \(-286.0\) | \(–394.0\) | \(–1560.0\) |
1. \(+54.0~\text{kJ mol}^{–1} \)
2. \(-68.0~\text{kJ mol}^{–1} \)
3. \(-86.0~\text{kJ mol}^{–1} \)
4. \(+97.0~\text{kJ mol}^{–1} \)
Subtopic: Hess's Law | Thermochemistry |
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At \(25^\circ\text{C}\) and 1 atm pressure, the enthalpies of combustion of benzene(l) and acetylene(g) are \(-3268~\text{kJ mol}^{-1}\) and \(-1300~\text{kJ mol}^{-1},\) respectively. Find out the change in enthalpy for the reaction \(3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l}),\) :
| 1. | \(+324~\text{kJ mol}^{-1}\) | 2. | \(+632~\text{kJ mol}^{-1}\) |
| 3. | \(-632~\text{kJ mol}^{-1}\) | 4. | \(-732~\text{kJ mol}^{-1}\) |
Subtopic: Hess's Law |
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\(\mathrm{17.0 ~g}\) of \(\mathrm{NH_3}\) completely vapourises at \(\mathrm{– 33.42^\circ C}\) and \(\mathrm{1~ bar}\) pressure and the enthalpy change in the process is \(\mathrm{23.4 ~kJ ~mol^{–1} .}\) The enthalpy change for the vapourisation of \(\mathrm{85 ~g}\) of \(\mathrm{NH_3}\) under the same conditions is:
1. 427kJ
2. 117kJ
3. 213kJ
4. 117J
1. 427kJ
2. 117kJ
3. 213kJ
4. 117J
Subtopic: Hess's Law | Thermochemistry |
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Consider the following reactions:
\(\begin{aligned} & \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+400 \mathrm{~kJ} \\ & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+100 \mathrm{~kJ} \end{aligned}\)
When coal of 60% purity is allowed to burn in the presence of insufficient oxygen, 60% of carbon is converted into ‘CO’ and the remaining is converted into ‘CO2’. Then, how much heat is generated when 0.6 kg of coal is burnt?
1. 1600 kJ
2. 3200 kJ
3. 4400 kJ
4. 6600 kJ
\(\begin{aligned} & \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+400 \mathrm{~kJ} \\ & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+100 \mathrm{~kJ} \end{aligned}\)
When coal of 60% purity is allowed to burn in the presence of insufficient oxygen, 60% of carbon is converted into ‘CO’ and the remaining is converted into ‘CO2’. Then, how much heat is generated when 0.6 kg of coal is burnt?
1. 1600 kJ
2. 3200 kJ
3. 4400 kJ
4. 6600 kJ
Subtopic: Hess's Law |
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Given the standard heats of combustion of ethane, hydrogen, and graphite as -1560 kJ/mol, -393.5 kJ/mol, and -286 kJ/mol respectively, calculate the standard heat of formation (\(Δ_fH^\circ_{298}\)) of ethane in kJ/mol:
1. \(-112.5 \)
2. \(-92.5 \)
3. \(-192.5 \)
4. \(-350 \)
1. \(-112.5 \)
2. \(-92.5 \)
3. \(-192.5 \)
4. \(-350 \)
Subtopic: Hess's Law | Thermochemistry |
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Identify the correct relation between x and y.
The enthalpy required to convert Br₂(l) into gaseous bromine atoms is x kJ mol⁻¹, while the bond enthalpy of Br–Br bond in Br₂(g) is y kJ mol⁻¹.
1. x and y are equal (x = y)
2. x is less than y (x < y)
3. x and y are not related
4. x is greater than y ( x > y)
Subtopic: Hess's Law | Thermochemistry |
Level 3: 35%-60%
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Given
\(\begin{aligned} &\mathrm{{C}_{{(graphite) }}+{O}_{2}({~g})} → \mathrm{{CO}_{2}({~g})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=-393.5 {~kJ} {~mol}^{-1}} \\ &\mathrm{H_{2}(g) + \frac{1}{2} {O}_{2}({~g})} → \mathrm{{H}_{2} {O}({l})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=-285.8 {~kJ} {~mol}^{-1}} \\ &\mathrm{{CO}_{2}({~g})+2 {H}_{2} {O}({l})} → \mathrm{{CH}_{4}({~g})+2 {O}_{2}({~g})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=+890.3 {~kJ} {~mol}^{-1}} \end{aligned}\)
Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction
\(\mathrm{C_{(graphite)} + 2 H_{2} (g) → CH_{4} (g)}\) will be :
1. –74.8 kJ mol–1
2. –144.0 kJ mol–1
3. +74.8 kJ mol–1
4. +144.0 kJ mol–1
Subtopic: Hess's Law |
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