Correct answer Carries: 4.
Wrong Answer Carries: -1.
For the reaction \( C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l) \), \( \Delta H = -1299.5 \, \text{kJ/mol} \) at 298 K. What is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))
For \( \Delta n_g = 2 - (1 + 2.5) = -1.5 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -1299.50 - (-1.5 \times 2.4776) = -1299.50 + 3.7164 = -1295.78 \, \text{kJ/mol} \).
Which of the following represents an extensive property?
Extensive properties depend on the amount of substance. Enthalpy (\(H\)) varies with the system’s size, unlike pressure, temperature, or density.
For an ideal gas, if \(C_v = 12.5 \, \text{J/mol·K}\), what is \(C_p\)? (\(R = 8.314 \, \text{J/mol·K}\))
For an ideal gas, \(C_p = C_v + R = 12.5 + 8.314 = 20.814 \, \text{J/mol·K}\).
Calculate the work done when 2 mol of an ideal gas expands adiabatically and reversibly from 10 L to 20 L at 600 K (\(\gamma = 1.4\)). (\(R = 8.314 \, \text{J/mol·K}\))
\(T_2 = 600 \times (10/20)^{0.4} = 600 \times 0.7579 = 454.74 \, \text{K}\), \(C_v = R/(\gamma-1) = 8.314 / 0.4 = 20.785 \, \text{J/mol·K}\), \(w = nC_v\Delta T = 2 \times 20.785 \times (454.74 - 600) = -6042 \, \text{J}\).
Calculate \(\Delta H\) for \(2C(s) + 3H_2(g) \rightarrow C_2H_6(g)\) if \(\Delta H_f^\circ (C_2H_6,g) = -84.7 \, \text{kJ/mol}\).
\(\Delta H = \Delta H_f^\circ (C_2H_6) - [2 \times \Delta H_f^\circ (C) + 3 \times \Delta H_f^\circ (H_2)] = -84.7 - [0 + 0] = -84.7 \, \text{kJ/mol}\).
Calculate \(\Delta H\) for \(C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)\) given: \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -283 \, \text{kJ/mol}\); \(H_2O(g) \rightarrow H_2(g) + \frac{1}{2}O_2(g)\), \(\Delta H = 241.8 \, \text{kJ/mol}\).
Reverse second: \(\Delta H = 283 \, \text{kJ}\). Add third: \(\Delta H = 241.8 \, \text{kJ}\). Subtract first: \(\Delta H = 283 + 241.8 - 393.5 = 131.3 \, \text{kJ/mol}\).
A reaction has \(\Delta H = 60 \, \text{kJ/mol}\) and \(\Delta S = 150 \, \text{J/K}\). At what temperature does \(\Delta G = 0\)?
\(\Delta G = \Delta H - T\Delta S = 0\), \(60 \times 10^3 - T \times 150 = 0\), \(T = 60000 / 150 = 400 \, \text{K}\).
For a reaction with \(\Delta H = -120 \, \text{kJ}\) and \(\Delta G = -130 \, \text{kJ}\) at 298 K, what is \(\Delta S\)?
\(\Delta G = \Delta H - T\Delta S\), \(-130 = -120 - 298 \Delta S\), \(\Delta S = (-130 + 120) / -298 = 0.0336 \, \text{kJ/K} = 33.6 \, \text{J/K}\).
Calculate \(\Delta H\) for \(C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\) using: \(\Delta H_f^\circ (C_3H_8,g) = -103.8 \, \text{kJ/mol}\), \(\Delta H_f^\circ (CO_2,g) = -393.5 \, \text{kJ/mol}\), \(\Delta H_f^\circ (H_2O,l) = -285.8 \, \text{kJ/mol}\).
\(\Delta H = [3(-393.5) + 4(-285.8)] - [-103.8 + 5(0)] = -1180.5 - 1143.2 + 103.8 = -2219.9 \, \text{kJ/mol}\).
Which of the following is true for an endothermic process with \(\Delta S > 0\)?
For an endothermic process (\(\Delta H > 0\)) with \(\Delta S > 0\), \(\Delta G = \Delta H - T\Delta S\). It becomes spontaneous (\(\Delta G < 0\)) at high temperatures when \(T\Delta S > \Delta H\).
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