At a certain temperature of 913 K, [tex]K_p[/tex] for the reaction [tex]2 \text{Cl}(g) \rightleftharpoons \text{Cl}_2(g)[/tex] is [tex]6.32 \times 10^{29}[/tex]. Calculate the value of [tex]\Delta G^\circ[/tex] in kJ for the reaction at 913 K.

To calculate the value of ΔG° in kJ for the reaction 2Cl(g) ⇌ Cl_{2} (g) at 913 K, given that the equilibrium constant, Kp, is 6.32 x [tex]10^{29}[/tex], we can follow these steps:

Step 1: Use the formula ΔG° = -RT ln(Kp) to calculate ΔG°, where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Kp is the equilibrium constant.

Step 2: Convert R to kJ/mol·K by dividing by 1000, so R = 0.008314 kJ/mol·K.

Step 3: Plug in the values into the formula: ΔG° = - (0.008314 kJ/mol·K) × (913 K) × ln(6.32 x [tex]10^{29}[/tex]).

Step 4: Calculate ΔG°, which equals - (0.008314 kJ/mol·K) × (913 K) × ln(6.32 x [tex]10^{29}[/tex]) ≈ -161.4 kJ/mol.

Therefore, the value of ΔG° for the reaction 2Cl(g) ⇌Cl_{2} (g) at 913 K is approximately -161.4 kJ/mol.

Note that the negative sign indicates that the reaction is spontaneous in the forward direction at this temperature.

To know more about Gibbs Energy:

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