In the reaction $\ce{E(g) <=> P(g)}$ at 25 °C, equilibrium is reached when the pressure of the product is 0.100 that of the reactant. What is $\Delta G$ in joules?

Using $\Delta G = RT\ln K$, my answer is $-8.31 \times 298 \times \ln (1.00) \ \text{J}$. However, the answer is $-8.31 \times 298 \times \ln (0.100) \ \text{J}$. Wouldn't this mean that $\Delta G > 0$, therefore favoring the reactants? (As an aside, why would the reaction even be shifted to the left since it has a higher pressure?)

  • $\begingroup$ You need to look again. $\Delta G$ is negative. $\endgroup$
    – LDC3
    Commented Apr 3, 2014 at 2:30
  • $\begingroup$ Wouldn't the negative value from ln(0.100) cancel with the negative in front? $\endgroup$
    – halcyon
    Commented Apr 3, 2014 at 2:31
  • $\begingroup$ Sorry, I didn't notice you changed from 1.0 to 0.1. But as it is stated, pressure of the products = 0.1 times the pressure of the reactants, which means there are more reactants. $\endgroup$
    – LDC3
    Commented Apr 3, 2014 at 3:08
  • $\begingroup$ Okay, that part makes sense. But then how does that correlate with the system being in equilibrium? $\endgroup$
    – halcyon
    Commented Apr 3, 2014 at 3:11
  • 1
    $\begingroup$ The equation is different, the web page has $\Delta G° = \Delta G + RT$ ln $K$. $\Delta G$ is 0 at equilibrium, so $\Delta G°$ is $-RT$ ln $K$ or −8.31∗298∗ln(0.100) J. $\endgroup$
    – LDC3
    Commented Apr 3, 2014 at 3:52

1 Answer 1


You are mixing up two different quantities. The equation you are using describes the standard Gibbs free energy of reaction $\Delta G^{0}$, i.e. $\Delta G$ for standard conditions without any requirement of the reaction being in equilibrium. And this quantity can very well be different from zero. For non-standard conditions the Gibbs free energy of reaction is

\begin{equation} \Delta G = \Delta G^{0} + RT \log K \end{equation}

and as you stated correctly this quantity must (per definition) be equal to zero for equilibrium conditions.

For a derivation see this answer of mine.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.