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  4. Chemical Thermodynamics Chapter-Wise Test 3

Chemical Thermodynamics Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Calculate \(\Delta H\) for \(C(s) + H_2(g) + O_2(g) \rightarrow CH_3OH(l)\) given: \(CH_3OH(l) + \frac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\), \(\Delta H = -726.5 \, \text{kJ/mol}\); \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\), \(\Delta H = -285.8 \, \text{kJ/mol}\).

Reverse first: \(\Delta H = 726.5 \, \text{kJ}\). Second: \(\Delta H = -393.5 \, \text{kJ}\). Third (×2): \(\Delta H = -571.6 \, \text{kJ}\). Add: \(726.5 - 393.5 - 571.6 = -238.6 \, \text{kJ/mol}\).

238.6 kJ/mol
-238.6 kJ/mol
-726.5 kJ/mol
679.3 kJ/mol
2

Calculate the bond enthalpy of \(C-C\) in \(C_2H_6(g)\) given: \(\Delta H_f^\circ (C_2H_6,g) = -84.7 \, \text{kJ/mol}\), \(\Delta H_a (C,g) = 715 \, \text{kJ/mol}\), \(\Delta H_a (H,g) = 218 \, \text{kJ/mol}\), \(C-H = 413 \, \text{kJ/mol}\).

For \(C_2H_6(g) \rightarrow 2C(g) + 6H(g)\), \(\Delta H = 2 \times 715 + 6 \times 218 - (-84.7) = 1430 + 1308 + 84.7 = 2822.7 \, \text{kJ/mol}\). Bonds: \(1(C-C) + 6(C-H)\), \(C-C = 2822.7 - 6 \times 413 = 2822.7 - 2478 = 344.7 \, \text{kJ/mol}\).

413 kJ/mol
715 kJ/mol
344.7 kJ/mol
2822.7 kJ/mol
3

Calculate \( \Delta S_{surr} \) for the combustion of 1 mol of \( CH_4(g) \) at 298 K, given \( \Delta H = -890.00 \, \text{kJ/mol} \).

For the surroundings, \( \Delta S_{surr} = -\Delta H / T = -(-890.00 \times 10^3) / 298 \approx 2986.6 \, \text{J/K} \).

-2986.6 J/K
890 J/K
-890 J/K
2986.6 J/K
4

Calculate \(\Delta S_{surr}\) when 3 mol of a liquid freezes at 250 K if \(\Delta H_{fus} = 5.5 \, \text{kJ/mol}\).

For freezing, \(\Delta H = -n \times \Delta H_{fus} = -3 \times 5.5 = -16.5 \, \text{kJ}\), \(\Delta S_{surr} = -\Delta H / T = -(-16.5 \times 10^3) / 250 = 66 \, \text{J/K}\).

-66 J/K
5.5 J/K
66 J/K
22 J/K
3

A system releases 200 J of heat and has 150 J of work done on it. What is \(\Delta U\)?

\(\Delta U = q + w\). Here, \(q = -200 \, \text{J}\) (heat released), \(w = +150 \, \text{J}\) (work done on system), so \(\Delta U = -200 + 150 = -50 \, \text{J}\).

350 J
-50 J
50 J
-350 J
2

The enthalpy change for the process \(H_2O(s) \rightarrow H_2O(l)\) is:

The enthalpy change for melting is the enthalpy of fusion (\(\Delta H_{fus}\)), which is positive as it’s an endothermic process.

\(\Delta H_{vap}\)
\(\Delta H_{fus}\)
\(-\Delta H_{fus}\)
\(\Delta H_{comb}\)
2

The enthalpy of formation of \(NO_2(g)\) is 33.2 kJ/mol. What is \(\Delta H\) for \(N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)\)?

\(\Delta H = 2 \times \Delta H_f^\circ (NO_2) - [\Delta H_f^\circ (N_2) + 2 \times \Delta H_f^\circ (O_2)] = 2 \times 33.2 - [0 + 0] = 66.4 \, \text{kJ/mol}\).

66.4 kJ/mol
33.2 kJ/mol
-66.4 kJ/mol
132.8 kJ/mol
1

In a reversible isothermal compression of 2 mol of an ideal gas at 300 K, the work done is 4988 J. What is the heat absorbed? (\(R = 8.314 \, \text{J/mol·K}\))

For isothermal process, \(\Delta U = 0\), so \(q = -w\). Compression means \(w > 0\) (4988 J), so \(q = -4988 \, \text{J}\).

-4988 J
4988 J
2494 J
0 J
1

For an exothermic reaction with \( \Delta H = -100.00 \, \text{kJ/mol} \) and \( \Delta n_g = -1 \) at 298 K, what is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = -1 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -100.00 - (-1 \times 2.4776) = -100.00 + 2.4776 = -97.52 \, \text{kJ/mol} \).

-100 kJ
-97.5 kJ
-102.5 kJ
-95 kJ
2

For a reaction with \(\Delta H = -50 \, \text{kJ}\) and \(\Delta S = 100 \, \text{J/K}\), at what temperature is \(\Delta G = 0\)?

\(\Delta G = \Delta H - T\Delta S = 0\), so \(-50 \times 10^3 + T \times 100 = 0\), \(T = 50000 / 100 = 500 \, \text{K}\).

250 K
1000 K
500 K
50 K
3

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