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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?)

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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