The value of the equilibrium constant for a particular reaction is [tex]6.32 \times 10^6[/tex] at 298 K. Calculate [tex]\Delta G^\circ[/tex] for this reaction. (R = [tex]8.31 \times 10^{-3}[/tex] kJ/K)

A. -38.8 kJ
B. -2550 kJ
C. 128 kJ
D. -7.84 kJ
E. 2.48 kJ

To calculate the standard Gibbs free energy change (ΔGº) for a reaction, we can use the equation ΔGº = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin.

In this case, the given equilibrium constant (K) is 6.32 x 10^6 and the temperature (T) is 298 K. The gas constant (R) is 8.31 x 10^3 kJ/K.

Substituting these values into the equation, we have:

ΔGº = - (8.31 x 10^3 kJ/K) x (298 K) x ln(6.32 x 10^6)

Calculating this expression, we find that ΔGº is approximately -38.8 kJ.

Learn more about calculating agº for a reaction here:

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The value of the equilibrium constant for a particular reaction is [tex]6.32 \times 10^6[/tex] at 298 K. Calculate [tex]\Delta G^\circ[/tex] for this reaction. (R = [tex]8.31 \times 10^{-3}[/tex] kJ/K)A. -38.8 kJ B. -2550 kJ C. 128 kJ D. -7.84 kJ E. 2.48 kJ

Answer :

Final answer:

Explanation:

Other Questions

The value of the equilibrium constant for a particular reaction is [tex]6.32 \times 10^6[/tex] at 298 K. Calculate [tex]\Delta G^\circ[/tex] for this reaction. (R = [tex]8.31 \times 10^{-3}[/tex] kJ/K)

A. -38.8 kJ
B. -2550 kJ
C. 128 kJ
D. -7.84 kJ
E. 2.48 kJ