Thermodynamics Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For a reaction with \(K = 0.01\) at 298 K, calculate \(\Delta G^\circ\). (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta G^\circ = -RT \ln K = -8.314 \times 298 \times \ln(0.01) = -8.314 \times 298 \times (-4.605) = 11403 \, \text{J/mol} = 11.4 \, \text{kJ/mol}\).

-11.4 kJ/mol
5.7 kJ/mol
-5.7 kJ/mol
11.4 kJ/mol
4

For \( H_2O(l) \rightarrow H_2(g) + \frac{1}{2}O_2(g) \), \( \Delta H = 285.80 \, \text{kJ/mol} \) at 298 K. What is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 1.5 - 0 = 1.5 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = 285.80 - 1.5 \times 2.4776 = 285.80 - 3.7164 = 282.08 \, \text{kJ/mol} \).

285.8 kJ/mol
282.1 kJ/mol
289.5 kJ/mol
279.3 kJ/mol
2

For which process is \(\Delta H = \Delta U\)?

\(\Delta H = \Delta U + \Delta n_g RT\). When \(\Delta n_g = 0\) (no change in moles of gas), \(\Delta H = \Delta U\), as in \(H_2(g) + Cl_2(g) \rightarrow 2HCl(g)\).

\(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)\)
\(H_2(g) + Cl_2(g) \rightarrow 2HCl(g)\)
\(C(s) + O_2(g) \rightarrow CO_2(g)\)
\(N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\)
2

For a reaction with \( \Delta H = -80.00 \, \text{kJ/mol} \) and \( \Delta S = -200.0 \, \text{J/K·mol} \), what is \( \Delta G \) at 300 K?

\( \Delta G = \Delta H - T\Delta S = -80.00 - 300 \times (-0.2000) = -80.00 + 60.00 = -20.00 \, \text{kJ/mol} \). The negative \( \Delta G \) indicates spontaneity at 300 K.

80 kJ/mol
-60 kJ/mol
20 kJ/mol
-20 kJ/mol
4

Calculate the heat absorbed when 27 g of aluminum is heated from 25°C to 125°C at constant volume. (\(C_v = 22.4 \, \text{J/mol·K}\), molar mass = 27 g/mol)

Moles = \(27 / 27 = 1 \, \text{mol}\), \(\Delta T = 125 - 25 = 100 \, \text{K}\), \(q_v = nC_v\Delta T = 1 \times 22.4 \times 100 = 2240 \, \text{J} = 2.24 \, \text{kJ}\).

1.12 kJ
4.48 kJ
22.4 kJ
2.24 kJ
4

For an ideal gas (\(\gamma = 1.5\)) expanding adiabatically from 10 L to 25 L at 600 K, what is the final temperature?

\(T_2 = T_1 (V_1/V_2)^{\gamma-1} = 600 \times (10/25)^{0.5} = 600 \times (0.4)^{0.5} = 600 \times 0.632 = 379.2 \, \text{K}\).

240 K
379.2 K
948 K
600 K
2

For an adiabatic process, which condition must be satisfied?

In an adiabatic process, no heat is exchanged between the system and surroundings, so \(q = 0\).

\(q = 0\)
\(\Delta T = 0\)
\(\Delta p = 0\)
\(w = 0\)
1

For the process \( 2NO_2(g) \rightarrow N_2O_4(g) \), \( \Delta H = -57.20 \, \text{kJ/mol} \) at 298 K, what is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 1 - 2 = -1 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -57.20 - (-1 \times 2.4776) = -57.20 + 2.4776 = -54.72 \, \text{kJ/mol} \).

-59.68 kJ/mol
-57.2 kJ/mol
-52.24 kJ/mol
-54.72 kJ/mol
4

Calculate \(\Delta S\) for the fusion of 36 g of ice at 273 K if \(\Delta H_{fus} = 6.01 \, \text{kJ/mol}\). (Molar mass of \(H_2O = 18 \, \text{g/mol}\))

Moles = \(36 / 18 = 2 \, \text{mol}\), \(\Delta S = n \times \Delta H_{fus} / T = 2 \times 6.01 \times 10^3 / 273 = 44 \, \text{J/K}\).

22 J/K
6.01 J/K
44 J/K
88 J/K
3

Calculate \(\Delta G^\circ\) for a reaction with \(K = 1.5 \times 10^{-4}\) at 298 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta G^\circ = -RT \ln K = -8.314 \times 298 \times \ln(1.5 \times 10^{-4}) = -2477.6 \times (-8.804) = 21815 \, \text{J/mol} = 21.82 \, \text{kJ/mol}\).

-21.82 kJ/mol
8.8 kJ/mol
-8.8 kJ/mol
21.82 kJ/mol
4

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