Question #7b856

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Question #7b856

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Answer 1 · 2015-01-26T13:15:17

Since this is an equilibrium reaction, the data given point to the formation of more reactants.

The Gibbs free energy change for a system that has a reaction quotient equal to $Q_c$ is

$DeltaG_("rxn") = DeltaG^@ + RT * ln(Q_c)$

You know that at equilibrium, $DeltaG = 0$ and $DeltaG^@ = -RT * ln(K_c)$ where $K_c$ is the equilibrium constant for the reaction.

The above equation can be written as

$DeltaG_("rxn")= - RT * ln(K_c) + RT * ln(Q_c)$ , or

$DeltaG_("rxn") = RT * ln(Q_c/K_c)$

This means that

$ln(Q_c/K_c) = (DeltaG_("rxn"))/(RT)$

Since $R$ and $T$ will always be positive, the value on the right of the equation is positive, which means that $(Q_c)/(K_c)$ is greater than 1, or that $Q_c$ is greater than $K_c$ .

This means that the reverse reaction will be spontaneous and the formation of more reactants will be favored.

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Question #7b856

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