# Question #dde3a

##### 1 Answer
Jan 26, 2015

So, you were given this equilibrium reaction

${A}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {B}_{\left(a q\right)} + {C}_{\left(a q\right)}$

Now, an equilibrium reaction is comprised of two reactions that take place at the same time - a forward reaction, which is the one given to you, and a reverse reaction, which looks like this

${B}_{\left(a q\right)} + {C}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {A}_{\left(a q\right)}$

This aspect is important because the forward reaction will be spontaneous if ${K}_{\text{forward}} > 1$; likewise, the reverse reaction will be spontaneous if ${K}_{\text{reverse") = 1/K_("forward}} > 1$.

A reaction's spontaneity is determined by the sign of its Gibbs free energy, $\Delta {G}^{\circ}$. If $\Delta {G}^{\circ}$ is negative for the forward reaction, then the forward reaction will be spontaneous. If $\Delta {G}^{\circ}$ is positive for the forward reaction, the forward reaction will not be spontaneous, but the reverse reaction will.

So, the equilibrium constant for the forward reaction is

${K}_{\text{forward}} = \frac{\left[B\right] \cdot \left[C\right]}{\left[A\right]} = \frac{0.0200 M \cdot 0.1 M}{\left[A\right]} = 0.100$

The expression for $\Delta {G}^{\circ}$ at equilibrium is

$\Delta {G}^{\circ} = - R T \cdot \ln \left(K\right)$

Notice that in order for $\Delta {G}^{\circ}$ to be negative, the natural log of $K$ must be positive, which atutomatically means that $K$ must be greater than 1.

Since no information about temperature is given, the equilibrium constant will remain unchanged during the reaction, which means that the reverse reaction will be spontaneous, since

${K}_{\text{rverse") = 1/K_("forward}} = \frac{1}{0.100} = 10.0$

This means that

${K}_{\text{reverse}} = \frac{\left[A\right]}{\left[B\right] \cdot \left[C\right]} = \frac{\left[A\right]}{0.0200 \cdot 0.1} = \frac{\left[A\right]}{0.002} = 10.0$

$\left[A\right] = 0.002 \cdot 10.0 = \text{0.02 mol/L}$, or

$\left[A\right] = \text{20 mmol/L}$

Now, don't forget about the reaction quotient, ${Q}_{c}$. If ${Q}_{c}$ is greater than ${K}_{\text{forward}}$, the reaction will shift to the left, favoring the formation of more reactant. SInce the reverse reaction is spontaneous, this means that you could get any value for the concentration of $A$ that satisfies this inequality

${Q}_{c} = \frac{0.002}{\left[A\right]} > 0.1 \implies \left[A\right] < 0.02 M$