Thermodynamics Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For the vaporization of 72 g of water at 373 K (\( \Delta H_{vap} = 40.79 \, \text{kJ/mol} \), molar mass = 18 g/mol), what is \( \Delta S_{sys} \)?

Moles = \( 72 / 18 = 4 \, \text{mol} \). For vaporization, \( \Delta S_{sys} = n \times \Delta H_{vap} / T = 4 \times 40790 / 373 \approx 437.48 \, \text{J/K} \approx 437.5 \, \text{J/K} \).

109.4 J/K
437.5 J/K
218.8 J/K
40.79 J/K
2

For \( CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g) \), \( \Delta H = -103.00 \, \text{kJ/mol} \) at 298 K. What is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 2 - 2 = 0 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -103.00 - 0 = -103.00 \, \text{kJ/mol} \).

-105.5 kJ/mol
-103 kJ/mol
-100.5 kJ/mol
-98 kJ/mol
2

For a reaction, \(\Delta H = -150 \, \text{kJ}\) and \(\Delta S = -0.5 \, \text{kJ/K}\). Below what temperature is it spontaneous?

\(\Delta G = \Delta H - T\Delta S < 0\), \(-150 - T(-0.5) < 0\), \(0.5T < 150\), \(T < 300 \, \text{K}\).

Above 300 K
Below 300 K
Above 600 K
Below 150 K
2

For the process \(N_2(g) \rightarrow 2N(g)\), what are the signs of \(\Delta H\) and \(\Delta S\)?

Bond breaking is endothermic (\(\Delta H > 0\)), and entropy increases as one molecule becomes two (\(\Delta S > 0\)).

\(\Delta H < 0, \Delta S < 0\)
\(\Delta H < 0, \Delta S > 0\)
\(\Delta H > 0, \Delta S < 0\)
\(\Delta H > 0, \Delta S > 0\)
4

Which property is intensive?

Intensive properties do not depend on the amount of substance. Temperature depends on the system’s state, not its size, unlike volume or enthalpy.

Volume
Temperature
Enthalpy
Mass
2

For \( 2SO_3(g) \rightarrow 2SO_2(g) + O_2(g) \), \( \Delta H = 197.80 \, \text{kJ/mol} \) at 298 K, what is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 3 - 2 = 1 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = 197.80 - 1 \times 2.4776 = 195.32 \, \text{kJ/mol} \).

200.3 kJ/mol
197.8 kJ/mol
2.48 kJ/mol
195.3 kJ/mol
4

Calculate \(\Delta H\) for \(C_2H_2(g) + 2H_2(g) \rightarrow C_2H_6(g)\) using: \(\Delta H_f^\circ (C_2H_2,g) = 226.7 \, \text{kJ/mol}\), \(\Delta H_f^\circ (C_2H_6,g) = -84.7 \, \text{kJ/mol}\).

\(\Delta H = \Delta H_f^\circ (C_2H_6) - [\Delta H_f^\circ (C_2H_2) + 2 \times \Delta H_f^\circ (H_2)] = -84.7 - [226.7 + 0] = -84.7 - 226.7 = -311.4 \, \text{kJ/mol}\).

311.4 kJ/mol
142 kJ/mol
-311.4 kJ/mol
-226.7 kJ/mol
3

The enthalpy of vaporization of a liquid is 32.00 kJ/mol at 350 K. What is \( \Delta U \) if \( \Delta n_g = 1 \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 1 \), \( RT = 8.314 \times 350 \times 10^{-3} = 2.9099 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = 32.00 - 2.9099 = 29.09 \, \text{kJ/mol} \).

32 kJ/mol
29.1 kJ/mol
34.9 kJ/mol
27 kJ/mol
2

For an ideal gas (\( \gamma = 1.4 \)) expanding adiabatically from 5 L to 10 L at 500 K, calculate the work done if \( C_v = 20.785 \, \text{J/mol·K} \). (1 mol gas)

For adiabatic expansion, \( T_2 = 500 \times (5/10)^{0.4} \approx 500 \times 0.7579 = 378.9 \, \text{K} \). Then, \( \Delta T = 378.9 - 500 = -121.1 \, \text{K} \). Work done: \( w = nC_v\Delta T = 1 \times 20.785 \times (-121.1) \approx -2516 \, \text{J} \).

2516 J
-1258 J
-2516 J
0 J
3

Calculate the work done when 1 mol of an ideal gas expands irreversibly against a constant external pressure of 2 atm from 5 L to 10 L. (1 atm·L = 101.3 J)

\(w = -P_{ext} \Delta V = -2 \times (10 - 5) \times 101.3 = -2 \times 5 \times 101.3 = -1013 \, \text{J}\).

-1013 J
1013 J
-506.5 J
2026 J
1

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