Question

The standard Gibbs free energy change for the reaction:
$2NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+B{r}_2\left(g\right)$
is $+12.50\text{kJ/mole}$ at ${27}^{\circ }\text{C}$. What is the equilibrium constant for the reaction at ${27}^{\circ }\text{C}$?

• A. ${10}^5$
• B. ${10}^{-5}$
• C. $e^5$
• D. $e^{-5}$

Answer 1

The correct option is D $e^{-5}$

$\Delta G=-R\times T\times lnK$
$12.5\times 1000=-8.314\times 300\times lnK$
$lnK=5K=e^{-5}$