Question:
The standard Gibbs free energy change for the reaction
$H_2O (g) → H_2 (g) + ½ O_2(g)$ at 2500 K is $+118 kJ mol^{-1}$
The equilibrium constant for the reaction is
[Given: $R= 8.314 J K^{-1} mol^{-1}$]
The standard Gibbs free energy change for the reaction
$H_2O (g) → H_2 (g) + ½ O_2(g)$ at 2500 K is $+118 kJ mol^{-1}$
The equilibrium constant for the reaction is
[Given: $R= 8.314 J K^{-1} mol^{-1}$]
$H_2O (g) → H_2 (g) + ½ O_2(g)$ at 2500 K is $+118 kJ mol^{-1}$
The equilibrium constant for the reaction is
[Given: $R= 8.314 J K^{-1} mol^{-1}$]
Updated On: Oct 1, 2024
- 0.994
- 1.006
- 3.42 x $10^{-3}$
- 292.12
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The Correct Option is C
Solution and Explanation
The correct option is (C): 3.42 x $10^{-3}$
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