Thermodynamics Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Calculate \(\Delta S_{total}\) when 2 mol of water vapor condenses at 373 K if \(\Delta H_{vap} = 40.79 \, \text{kJ/mol}\).

\(\Delta H = -2 \times 40.79 = -81.58 \, \text{kJ}\), \(\Delta S_{sys} = -\Delta H_{vap} \times n / T = -40.79 \times 10^3 \times 2 / 373 = -218.8 \, \text{J/K}\), \(\Delta S_{surr} = -\Delta H / T = 81.58 \times 10^3 / 373 = 218.8 \, \text{J/K}\), \(\Delta S_{total} = -218.8 + 218.8 = 0 \, \text{J/K}\).

218.8 J/K
-218.8 J/K
0 J/K
437.6 J/K
3

Which of the following statements is true for a reversible adiabatic expansion of an ideal gas?

In a reversible adiabatic process (\(q = 0\)), \(\Delta U = w\). For an ideal gas, \(\Delta U = nC_v\Delta T\), and since expansion cools the gas, \(\Delta T < 0\), making \(\Delta U < 0\).

\(\Delta U < 0\)
\(\Delta H = 0\)
\(\Delta S > 0\)
\(q > 0\)
1

For an ideal gas (\( \gamma = 1.67 \)) compressed adiabatically from \(V_1 = 12\,\text{L}\) to \(V_2 = 3\,\text{L}\), starting at \(T_1 = 300\,\text{K}\), calculate the work done on the gas if \( C_v = 12.47 \, \text{J mol}^{-1}\text{K}^{-1} \). (Assume \(n=1\) mol.)

For an adiabatic ideal-gas process, \( T_2 = T_1 (V_1/V_2)^{\gamma-1} = 300 \times 4^{0.67} \approx 300 \times 2.5315 = 759.45 \,\text{K} \).

With \( q=0 \), \( w = \Delta U = n C_v (T_2 - T_1) = 1 \times 12.47 \times (759.45 - 300) \approx 5729 \,\text{J} \).

Since the work is done on the gas, it is positive: \( w \approx +5729 \,\text{J} \).

5729 J
-5729 J
3049 J
0 J
1

For \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), \( \Delta H = -92.20 \, \text{kJ/mol} \) at 298 K. What is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 2 - 4 = -2 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -92.20 - (-2 \times 2.4776) = -92.20 + 4.9552 = -87.24 \, \text{kJ/mol} \).

-87.2 kJ/mol
-92.2 kJ/mol
-97.2 kJ/mol
-82.3 kJ/mol
1

For an ideal gas expanding irreversibly against a constant external pressure of 1 atm from 2 L to 5 L, calculate the work done. (1 atm·L = 101.3 J)

\(w = -P_{ext} \Delta V = -1 \times (5 - 2) \times 101.3 = -1 \times 3 \times 101.3 = -303.9 \, \text{J}\).

-303.9 J
303.9 J
-607.8 J
101.3 J
1

Calculate the bond enthalpy of \( N≡N \) in \( N_2(g) \) given: \( \Delta H_f^\circ (NH_3,g) = -46.10 \, \text{kJ/mol} \), \( N-H = 391.0 \, \text{kJ/mol} \), \( \Delta H_a (H,g) = 218.0 \, \text{kJ/mol} \).

For \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), \( \Delta H = 2 \times (-46.10) = -92.20 \, \text{kJ} \). Bonds broken: \( N≡N + 3 \times H-H = N≡N + 3 \times 436.0 = N≡N + 1308.0 \, \text{kJ} \). Bonds formed: \( 6 \times N-H = 6 \times 391.0 = 2346.0 \, \text{kJ} \). Using \( \Delta H = \text{(bonds broken)} - \text{(bonds formed)} \), \( -92.20 = (N≡N + 1308.0) - 2346.0 \), \( N≡N = 945.8 \, \text{kJ/mol} \).

436 kJ/mol
391 kJ/mol
945.8 kJ/mol
2346 kJ/mol
3

Calculate the bond enthalpy of \(O=O\) in \(O_2(g)\) given: \(\Delta H_f^\circ (H_2O,g) = -241.8 \, \text{kJ/mol}\), \(H-H = 436 \, \text{kJ/mol}\), \(O-H = 463 \, \text{kJ/mol}\).

For \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)\), \(\Delta H = -241.8 \, \text{kJ}\). Bonds broken: \(H-H (436) + \frac{1}{2}O=O\). Bonds formed: \(2(O-H) = 2 \times 463 = 926\). \(\Delta H = (436 + \frac{1}{2}O=O) - 926 = -241.8\), \(\frac{1}{2}O=O = 248.2\), \(O=O = 496.4 \, \text{kJ/mol}\).

436 kJ/mol
926 kJ/mol
496.4 kJ/mol
241.8 kJ/mol
3

For the reaction \(N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)\), \(\Delta H = 66.4 \, \text{kJ/mol}\) at 298 K, what is \(\Delta U\)? (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta n_g = 2 - 3 = -1\), \(RT = 2.48 \, \text{kJ}\), \(\Delta H = \Delta U + \Delta n_g RT\), \(\Delta U = 66.4 - (-1 \times 2.48) = 66.4 + 2.48 = 68.88 \, \text{kJ/mol}\).

63.92 kJ/mol
66.4 kJ/mol
2.48 kJ/mol
68.88 kJ/mol
4

The heat released when 12 g of carbon is burnt in excess oxygen is: (\(\Delta H_c = -393.5 \, \text{kJ/mol}\), molar mass of \(C = 12 \, \text{g/mol}\))

Moles of \(C = 12 / 12 = 1 \, \text{mol}\). Heat released = \(1 \times 393.5 = 393.5 \, \text{kJ}\).

196.75 kJ
393.5 kJ
787 kJ
98.4 kJ
2

A reaction has \(\Delta H = -200 \, \text{kJ/mol}\) and \(\Delta S = -250 \, \text{J/K}\). At what temperature does it cease to be spontaneous?

\(\Delta G = 0\) at equilibrium, \(\Delta H - T\Delta S = 0\), \(-200 \times 10^3 - T(-250) = 0\), \(T = 200000 / 250 = 800 \, \text{K}\). Spontaneous below 800 K.

400 K
800 K
1000 K
200 K
2

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