Thermodynamics Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

250 K
1000 K
500 K
50 K
3

The heat capacity of a gas at constant volume is 20 J/mol·K. What is \(C_p\) if it is an ideal gas? (\(R = 8.314 \, \text{J/mol·K}\))

For an ideal gas, \(C_p = C_v + R = 20 + 8.314 = 28.314 \, \text{J/mol·K}\).

11.686 J/mol·K
20 J/mol·K
8.314 J/mol·K
28.3 J/mol·K
4

For an ideal gas undergoing an isochoric process, which of the following is always zero?

In an isochoric process (\(\Delta V = 0\)), work done \(w = -P\Delta V = 0\), regardless of other conditions.

Work done
Heat absorbed
\(\Delta U\)
\(\Delta H\)
1

For \( C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) \), \( \Delta H = -137.00 \, \text{kJ/mol} \) 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 \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -137.00 - (-1 \times 2.4776) = -137.00 + 2.4776 = -134.52 \, \text{kJ/mol} \).

-137 kJ/mol
-134.5 kJ/mol
-139.5 kJ/mol
-132 kJ/mol
2

A system absorbs 600 J of heat at constant volume, and its internal energy increases by 600 J. What is the work done?

At constant volume, \(w = -P\Delta V = 0\). First Law: \(\Delta U = q + w = 600 + 0 = 600 \, \text{J}\), so \(w = 0 \, \text{J}\).

0 J
600 J
-600 J
1200 J
1

For a reaction at constant volume, the heat absorbed is equal to:

At constant volume, \(w = -p\Delta V = 0\), so from the first law, \(\Delta U = q_v\), the heat absorbed at constant volume.

\(\Delta U\)
\(\Delta H\)
\(\Delta G\)
\(-\Delta H\)
1

The standard enthalpy of formation of \(CO_2(g)\) is -393.5 kJ/mol. This value refers to the formation of:

The standard enthalpy of formation is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states, i.e., \(C(s) + O_2(g) \rightarrow CO_2(g)\).

1 mol of \(CO_2(g)\) from \(C(s)\) and \(O_2(g)\)
2 mol of \(CO_2(g)\) from \(C(s)\) and \(O_2(g)\)
1 mol of \(CO_2(g)\) from \(CO(g)\) and \(O_2(g)\)
1 mol of \(CO_2(g)\) from \(C(g)\) and \(O_2(g)\)
1

Calculate \(\Delta G^\circ\) for a reaction with \(K = 2 \times 10^6\) at 298 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta G^\circ = -RT \ln K = -8.314 \times 298 \times \ln(2 \times 10^6) = -2477.6 \times 14.509 = -35962 \, \text{J/mol} = -35.96 \, \text{kJ/mol}\).

35.96 kJ/mol
14.51 kJ/mol
-14.51 kJ/mol
-35.96 kJ/mol
4

Calculate \(\Delta H\) for \(C(s) + 2S(s) \rightarrow CS_2(l)\) given: \(CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\), \(\Delta H = -1076.8 \, \text{kJ/mol}\); \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(S(s) + O_2(g) \rightarrow SO_2(g)\), \(\Delta H = -296.8 \, \text{kJ/mol}\).

Reverse first: \(\Delta H = 1076.8 \, \text{kJ}\). Second: \(\Delta H = -393.5 \, \text{kJ}\). Third (×2): \(\Delta H = -593.6 \, \text{kJ}\). Add: \(1076.8 - 393.5 - 593.6 = 89.7 \, \text{kJ/mol}\).

-89.7 kJ/mol
89.7 kJ/mol
1076.8 kJ/mol
-787 kJ/mol
2

For \( CO(g) + H_2O(g) \rightarrow CO_2(g) + H_2(g) \), \( \Delta H = -41.20 \, \text{kJ/mol} \) at 298 K, what is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = 2 - 2 = 0 \), \( \Delta H = \Delta U + \Delta n_g RT \). Since \( \Delta n_g = 0 \), \( \Delta U = \Delta H = -41.20 \, \text{kJ/mol} \).

-43.7 kJ/mol
41.2 kJ/mol
-38.7 kJ/mol
-41.2 kJ/mol
4

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