Question

A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.

Answer 1

Answer:

$\Delta G=-97.14kJ$

Explanation:

Hello,

In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

$\Delta G=-RTln(K)$

Hence, we compute it as required:

$\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol$

And for 2.37 moles of hydrogen bromide, we obtain:

$\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ$

Best regards.