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  4. Chemical Thermodynamics Chapter-Wise Test 2

Chemical Thermodynamics Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The enthalpy of combustion of benzene is -3267.6 kJ/mol. What is the heat released when 39 g of benzene is burnt? (Molar mass = 78 g/mol)

Moles = \(39 / 78 = 0.5 \, \text{mol}\), heat released = \(0.5 \times 3267.6 = 1633.8 \, \text{kJ}\).

1633.8 kJ
3267.6 kJ
816.9 kJ
6535.2 kJ
1

Using Hess’s law, calculate \(\Delta H\) for \(C(graphite) + 2H_2(g) \rightarrow CH_4(g)\), given: \(C(graphite) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\), \(\Delta H = -285.8 \, \text{kJ/mol}\); \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\), \(\Delta H = -890.3 \, \text{kJ/mol}\).

Reverse the third equation and add the first two: \(\Delta H = -393.5 + (-2 \times 285.8) + 890.3 = -74.8 \, \text{kJ/mol}\).

-52.3 kJ/mol
74.8 kJ/mol
-74.8 kJ/mol
-890.3 kJ/mol
3

Calculate \(\Delta S_{sys}\) when 1 mol of an ideal gas expands isothermally from 4 L to 16 L at 500 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta S = nR \ln(V_2/V_1) = 1 \times 8.314 \times \ln(16/4) = 8.314 \times \ln(4) = 8.314 \times 1.386 = 11.52 \, \text{J/K}\).

5.76 J/K
11.52 J/K
23.04 J/K
8.314 J/K
2

Calculate \(\Delta H\) for \(C(s) + CO_2(g) \rightarrow 2CO(g)\) using: \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -283 \, \text{kJ/mol}\).

The \(\Delta H_f\) for CO can be found from the second reaction: \(\Delta H_f (CO) = \Delta H_f (CO_2) + 283 = -393.5 + 283 = -110.5 \, \text{kJ/mol}\).

For the target reaction: \(\Delta H = 2 \times \Delta H_f (CO) - \Delta H_f (CO_2) = 2 \times (-110.5) - (-393.5) = -221 + 393.5 = 172.5 \, \text{kJ/mol}\).

-172.5 kJ/mol
172.5 kJ/mol
959.5 kJ/mol
-959.5 kJ/mol
2

For a reaction with \(\Delta H = 90 \, \text{kJ/mol}\) and \(\Delta G = 30 \, \text{kJ/mol}\) at 300 K, what is \(\Delta S\)?

\(\Delta G = \Delta H - T\Delta S\), \(30 = 90 - 300 \Delta S\), \(\Delta S = (90 - 30) / 300 = 60 / 300 = 0.2 \, \text{kJ/K} = 200 \, \text{J/K}\).

300 J/K
-200 J/K
200 J/K
90 J/K
3

Calculate \(\Delta S_{sys}\) when 1 mol of an ideal gas is compressed isothermally from 25 L to 5 L at 350 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta S = nR \ln(V_2/V_1) = 1 \times 8.314 \times \ln(5/25) = 8.314 \times \ln(0.2) = 8.314 \times (-1.609) = -13.38 \, \text{J/K}\).

13.38 J/K
-13.38 J/K
6.69 J/K
-26.76 J/K
2

Which of the following is a state function?

A state function depends only on the initial and final states, not the path taken. Internal energy (\(U\)) is a state function, while work (\(w\)) and heat (\(q\)) are path-dependent.

Internal energy
Work done
Heat absorbed
Rate of reaction
1

Calculate \( \Delta S_{sys} \) when 2 mol of an ideal gas expands isothermally from 1 atm to 0.25 atm at 300 K. (\( R = 8.314 \, \text{J/mol·K} \))

For isothermal expansion, \( \Delta S_{sys} = nR \ln(P_1/P_2) = 2 \times 8.314 \times \ln(1/0.25) = 16.628 \times 1.3863 \approx 23.0 \, \text{J/K} \).

11.5 J/K
16.6 J/K
23.0 J/K
46.1 J/K
3

Calculate the bond enthalpy of \( C=O \) in \( CO(g) \) given: \( \Delta H_f^\circ (CO,g) = -110.5 \, \text{kJ/mol} \), \( \Delta H_a (C,g) = 715.0 \, \text{kJ/mol} \), \( \Delta H_a (O,g) = 249.0 \, \text{kJ/mol} \).

For \( CO(g) \rightarrow C(g) + O(g) \), bond enthalpy is \( \Delta H = \Delta H_a(C) + \Delta H_a(O) - \Delta H_f(CO) = 715.0 + 249.0 - (-110.5) = 1074.5 \, \text{kJ/mol} \).

715 kJ/mol
964 kJ/mol
1074.5 kJ/mol
110.5 kJ/mol
3

Which of the following is always true for an exothermic process at constant pressure?

For an exothermic process, heat is released (\(q_p < 0\)), and at constant pressure, \(\Delta H = q_p\), so \(\Delta H < 0\).

\(\Delta H < 0\)
\(\Delta U < 0\)
\(\Delta S > 0\)
\(\Delta G < 0\)
1

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