Question

Using data given in table find out incorrect statement(s) among the following.

$CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$

Assume vibration modes of motion do not contribute to heat capacity at low temperature.

• A. $\Delta H^\circ > \Delta U^\circ$ for the reaction at 298K.
• B. In standard state condition, the reaction $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$ attain equilibrium at very high temperature.
• C. At low temperature $\frac{d(\Delta H)^\circ}{dT} = -ive$
• D. In a $CO, O_2$ fuel cell electrical energy obtained by cell $> |\Delta H^\circ_{combustion}[CO(g)]|$

Answer 1

Answer: (B), (C), (D)

A, D

Answer 2

The reaction is $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$.

(A) $\Delta H^\circ > \Delta U^\circ$ for the reaction at 298K. The relationship between $\Delta H^\circ$ and $\Delta U^\circ$ is $\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas. For the given reaction, $\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - (1 + \frac{1}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$. At 298K, R and T are positive. Thus, $\Delta n_g RT = -\frac{1}{2} RT$ is negative. $\Delta H^\circ - \Delta U^\circ = \Delta n_g RT < 0$, which means $\Delta H^\circ < \Delta U^\circ$. Therefore, the statement $\Delta H^\circ > \Delta U^\circ$ is incorrect.

(B) In standard state condition, the reaction $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$ attain equilibrium at very high temperature. For equilibrium, $\Delta G^\circ = 0$. $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$. The equilibrium temperature is $T_{eq} = \frac{\Delta H^\circ}{\Delta S^\circ}$. Using the given data: $\Delta H^\circ = \Delta H_f^\circ(CO_2) - [\Delta H_f^\circ(CO) + \frac{1}{2}\Delta H_f^\circ(O_2)] = -395 - [-110 + \frac{1}{2}(0)] = -395 + 110 = -285 \, \text{kJ/mol}$. $\Delta S^\circ = S^\circ(CO_2) - [S^\circ(CO) + \frac{1}{2}S^\circ(O_2)] = 213 - [197 + \frac{1}{2}(205)] = 213 - [197 + 102.5] = 213 - 299.5 = -86.5 \, \text{J/K mol} = -0.0865 \, \text{kJ/K mol}$. $T_{eq} = \frac{-285 \, \text{kJ/mol}}{-0.0865 \, \text{kJ/K mol}} \approx 3294.8 \, \text{K}$. Since the equilibrium temperature is very high, the statement is true.

(C) At low temperature $\frac{d(\Delta H)^\circ}{dT} = -ive$. According to Kirchhoff's law, $\frac{d(\Delta H)^\circ}{dT} = \Delta C_p^\circ$. At low temperature, assuming only translational and rotational contributions to heat capacity for ideal gases: For linear molecules (CO, $O_2$, $CO_2$), $C_p = C_V + R = (\frac{3}{2}R + R) + R = \frac{5}{2}R + R = \frac{7}{2}R$ (translational + rotational + R). Or, using degrees of freedom: $C_V = \frac{f}{2}R$, $C_p = C_V + R$. Translational degrees of freedom = 3 for all gases. Rotational degrees of freedom = 2 for linear molecules. $C_V(\text{linear}) = \frac{3}{2}R + \frac{2}{2}R = \frac{5}{2}R$. $C_p(\text{linear}) = \frac{5}{2}R + R = \frac{7}{2}R$. $\Delta C_p^\circ = C_p(CO_2) - [C_p(CO) + \frac{1}{2}C_p(O_2)] = \frac{7}{2}R - [\frac{7}{2}R + \frac{1}{2}(\frac{7}{2}R)] = \frac{7}{2}R - \frac{7}{2}R - \frac{7}{4}R = -\frac{7}{4}R$. Since R is positive, $\Delta C_p^\circ$ is negative. So, $\frac{d(\Delta H)^\circ}{dT} = \Delta C_p^\circ$ is negative. The statement is true.

(D) In a $CO, O_2$ fuel cell electrical energy obtained by cell $> |\Delta H^\circ_{combustion}[CO(g)]|$. The maximum electrical work obtained from a fuel cell operating reversibly at constant temperature and pressure is given by $-\Delta G^\circ$. The enthalpy of combustion of CO is $\Delta H^\circ_{combustion}[CO(g)] = \Delta H^\circ = -285 \, \text{kJ/mol}$. $|\Delta H^\circ_{combustion}[CO(g)]| = |-285| = 285 \, \text{kJ/mol}$. The electrical energy obtained is $-\Delta G^\circ = -(\Delta H^\circ - T \Delta S^\circ) = -\Delta H^\circ + T \Delta S^\circ$. We need to check if $-\Delta H^\circ + T \Delta S^\circ > -\Delta H^\circ$. This inequality simplifies to $T \Delta S^\circ > 0$. We calculated $\Delta S^\circ = -86.5 \, \text{J/K mol}$. At any positive temperature T, $T \Delta S^\circ$ will be negative. Since $T \Delta S^\circ < 0$, it means $-\Delta H^\circ + T \Delta S^\circ < -\Delta H^\circ$. So, the electrical energy obtained is less than the magnitude of the enthalpy of combustion. The statement is incorrect.

The incorrect statements are (A) and (D).