- 1(current)
Q. No.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 |
From NCERT
JEE
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Hints
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Given :
Based on the above thermochemical equations, the value of at 298 K for the reaction
1. 85.2 kJ/mol
2. 15.2 kJ/mol
3. -74.8 kJ/mol
4. 74.8 kJ/mol
Subtopic: Thermochemistry |
From NCERT
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Hints
The enthalpy of combustion of four allotropic forms of element 'X' are given as:
The most stable allotropic form of element 'X' is:
1. A
2. B
3. C
4. D
| Allotropic forms | \(\Delta_\text{comb}H^\circ\)(kJ/mol) |
| A. | -270.3 |
| B. | -189.1 |
| C. | -390.5 |
| D. | -465.0 |
1. A
2. B
3. C
4. D
Subtopic: Thermochemistry |
From NCERT
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Hints
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- 1(current)
Q. No.
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