Chapter 11: Nonisothermal Reactor Design: The Steady State Energy Balance and Adiabatic PFR Applications
Heats of Reaction
Calculate \( \Delta H_{rx}^\circ \), \( \Delta C_p \), and \( \Delta H_{rx} (400) \) for the reaction \( A \rightarrow 2B + C \)
There are inerts I present in the system.
Additional information
\( H_A^\circ(298) = -100 \, \text{kcal/mol} \, A \)
\( H_B^\circ(298) = -40 \, \text{kcal/mol} \, B \)
\( H_C^\circ(298) = -30 \, \text{kcal/mol} \, C \)
\( H_I^\circ(298) = -100 \, \text{kcal/mol} \, I \)
\( \tilde{C}_A = 80 \, \text{cal/mol} \, ^\circ C \)
\( \tilde{C}_B = 20 \, \text{cal/mol} \, ^\circ C \)
\( \tilde{C}_C = 30 \, \text{cal/mol} \, ^\circ C \)
\( \tilde{C}_I = 190 \, \text{cal/mol} \, ^\circ C \)
Hint 1: What is the heat of reaction at 298 C?
$(a) -10 \frac {cal}{mol A K}$
Calculate \( \Delta H_{rx}^\circ \), \( \Delta C_p \), and \( \Delta H_{rx} (400) \) for the reaction \( A \rightarrow 2B + C \)
Additional information
\( H_A^\circ(298) = -100 \, \text{kcal/mol} \, A \)
\( H_B^\circ(298) = -40 \, \text{kcal/mol} \, B \)
\( H_C^\circ(298) = -30 \, \text{kcal/mol} \, C \)
\( H_I^\circ(298) = -100 \, \text{kcal/mol} \, I \)
\( \tilde{C}_A = 80 \, \text{cal/mol} \, ^\circ C \)
\( \tilde{C}_B = 20 \, \text{cal/mol} \, ^\circ C \)
\( \tilde{C}_C = 30 \, \text{cal/mol} \, ^\circ C \)
\( \tilde{C}_I = 190 \, \text{cal/mol} \, ^\circ C \)
Solution
\( \Delta H_{Rx}^\circ(298) = H_C^\circ + 2H_B^\circ - H_A^\circ = (-30) + 2(-40) - (-100) = -10 \, \text{kcal/mol} \, A \)
\( \Delta \tilde{C}_P = \tilde{C}_{Pc} + 2 \tilde{C}_{Pb} - \tilde{C}_{Pa} = 30 + 2(20) - 80 = -10 \, \text{cal/mol} \, ^\circ C \)
\( \Delta H_{Rx}(T) = \Delta H_{Rx}^\circ(298) + \Delta C_P(T - T_R) \)
\( \Delta H_{Rx}(400) = -10 \, \text{kcal/mol} - 10 \, \text{cal/mol} \, \left( 400K - 298K \right) = -10,000 \, \text{cal/mol A} - (10)(102) \, \text{cal/mol} \)
\( = -11,200 \, \text{cal/mol A} \)
Note: The inerts NEVER enter into these calculations of \( \Delta H_{Rx}^\circ \) or \( \Delta \hat{C}_P \).