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------------------------------------------------ Evaluate [tex]\Delta G^0[/tex] at [tex]718^\circ \text{C}[/tex] for a gas phase reaction for which [tex]K_p = 7.4 \times 10^{-6}[/tex] at [tex]718^\circ \text{C}[/tex]. [tex]R = 8.314 \, \text{J/mol} \cdot \text{K}[/tex].

A. 68.6 kJ
B. 365 kJ
C. 427 kJ
D. 97.3 kJ
E. 168 kJ

Answer :

Final answer:

The ΔG° can be calculated using the equation ΔG° = -RT ln K, where K is the equilibrium constant. By substituting the given values into the equation, the ΔG° for the reaction can be determined as -96.8 kJ mol-¹. The correct answer is d. 97.3 kJ.

Explanation:

The question asks to evaluate δg0 at 718oc for a gas phase reaction for which kp = 7.4 x 10-6 at 718oc and r = 8.314 j/mol k.

To solve this, we can use the equation ΔG° = -RT ln K.

By substituting the given values into the equation, we can calculate the ΔG° for the reaction. ΔG° = (-8.314 J K mol-¹)(718 K)(ln(7.4 × 10-6)) = -96.8 kJ mol-¹.

Therefore, the answer is d. 97.3 kJ.

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