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

`\Delta G=-97.14kJ`

Step-by-step 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(J)/(mol* 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.