Correct answer Carries: 4.
Wrong Answer Carries: -1.
A reaction has \(\Delta H = -100 \, \text{kJ}\) and \(\Delta S = -200 \, \text{J/K}\). At what temperature does it become spontaneous?
For spontaneity, \(\Delta G = \Delta H - T\Delta S < 0\). So, \(-100 \times 10^3 - T(-200) < 0\), \(100000 > 200T\), \(T < 500 \, \text{K}\).
A reaction has \( \Delta H = 80.00 \, \text{kJ/mol} \), \( \Delta S = 200.0 \, \text{J/K·mol} \), and is spontaneous at 450 K. What is \( \Delta G \) at 450 K?
\( \Delta G = \Delta H - T\Delta S = 80.00 - 450 \times 0.2000 = 80.00 - 90.00 = -10.00 \, \text{kJ/mol} \). The negative \( \Delta G \) confirms the reaction is spontaneous at 450 K.
Calculate \(\Delta H\) for the reaction \(C(s) + 2H_2(g) \rightarrow CH_4(g)\) given: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\), \(\Delta H = -890.3 \, \text{kJ/mol}\); \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\), \(\Delta H = -285.8 \, \text{kJ/mol}\).
Reverse first: \(\Delta H = 890.3 \, \text{kJ}\). Second: \(\Delta H = -393.5 \, \text{kJ}\). Third (×2): \(\Delta H = -571.6 \, \text{kJ}\). Add: \(890.3 - 393.5 - 571.6 = -74.8 \, \text{kJ/mol}\).
For the process \(H_2O(l) \rightarrow H_2O(g)\) at 373 K, \(\Delta H = 40.79 \, \text{kJ/mol}\), what is \(\Delta S_{surr}\)?
\(\Delta S_{surr} = -\Delta H / T = -40.79 \times 10^3 / 373 = -109.4 \, \text{J/K·mol}\).
For the reaction \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \), \( \Delta H = -483.6 \, \text{kJ/mol} \) at 298 K. What is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))
For \( \Delta n_g = 2 - 3 = -1 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -483.60 - (-1 \times 2.4776) = -483.60 + 2.4776 = -481.12 \, \text{kJ/mol} \).
Which of the following is always true for a spontaneous process at constant temperature and pressure?
For a spontaneous process at constant \(T\) and \(P\), the Gibbs free energy must decrease (\(\Delta G < 0\)), as it determines spontaneity under these conditions.
The work done in the reversible adiabatic expansion of 1 mol of an ideal gas from 10 L to 20 L is (\( \gamma = 1.4 \)) at 300 K. Calculate \( w \). (\( R = 8.314 \, \text{J/mol·K} \))
For adiabatic reversible expansion, \( T_2 = T_1 (V_1/V_2)^{\gamma-1} = 300 \times (10/20)^{0.4} \approx 300 \times 0.7579 = 227.4 \, \text{K} \). Then, \( C_v = R/(\gamma-1) = 8.314 / 0.4 = 20.785 \, \text{J/mol·K} \). Work done: \( w = nC_v\Delta T = 1 \times 20.785 \times (227.4 - 300) \approx -1509.2 \, \text{J} \).
Calculate \(\Delta S_{surr}\) when 2 mol of a substance melts at 300 K if \(\Delta H_{fus} = 8 \, \text{kJ/mol}\).
\(\Delta H = 2 \times 8 = 16 \, \text{kJ}\), \(\Delta S_{surr} = -\Delta H / T = -16 \times 10^3 / 300 = -53.33 \, \text{J/K}\).
For an ideal gas expanding reversibly and adiabatically, the relationship between temperature and volume is:
For a reversible adiabatic process, \(TV^{\gamma-1} = \text{constant}\), where \(\gamma = C_p/C_v\).
The standard enthalpy of formation of \(NH_3(g)\) is -46.1 kJ/mol. What is \(\Delta H\) for \(N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\)?
\(\Delta H = 2 \times \Delta H_f^\circ (NH_3) - [\Delta H_f^\circ (N_2) + 3 \times \Delta H_f^\circ (H_2)] = 2 \times (-46.1) - [0 + 0] = -92.2 \, \text{kJ/mol}\).
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