# Question #5b2e7

Dec 8, 2015

(a)

$4.184$

(b)

$- 292 \text{kJ}$

#### Explanation:

(a)

${S}_{\left(s\right)} + {O}_{2 \left(g\right)} \rightarrow S {O}_{2 \left(g\right)}$

The expression to use which relates free energy change to the concentrations of all the reacting species is:

$\Delta {G}_{r} = \Delta {G}^{\circ} + R T \ln Q \text{ } \textcolor{red}{\left(1\right)}$

$Q$ is termed the reaction quotient which, in this case, is given by:

$Q = \frac{\left[S {O}_{2}\right]}{\left[{O}_{2}\right]}$

As the reaction proceeds to equilibrium the value of $\Delta {G}_{r}$ increases to zero. When this point is reached we can say from $\textcolor{red}{\left(1\right)}$ that:

$0 = \Delta {G}^{\circ} + R T \ln Q \text{ } \textcolor{red}{\left(2\right)}$

Because we are at equilibrium we can say that $Q = {K}_{c}$.

This means that $\textcolor{red}{\left(2\right)}$ becomes $\Rightarrow$

$\Delta {G}^{\circ} = - R T \ln {K}_{c} \text{ } \textcolor{red}{\left(3\right)}$

This is an important expression in chemistry as you can predict, given the relevant thermodynamic data, the equilibrium constant of a reaction.

This is what we can now do with this question:

From $\textcolor{red}{\left(3\right)}$ we get:

$\ln {K}_{c} = - \frac{\Delta {G}^{\circ}}{R T}$

$\therefore \ln {K}_{c} = - \left(\frac{- 300.4 \times {10}^{3}}{8.31 \times 298}\right)$

$\ln {K}_{c} = 121.3$

${K}_{c} = 4.814$

(b)

Now we just need to insert the values we have been given into eqn $\textcolor{red}{\left(1\right)} \Rightarrow$

$\Delta {G}_{r} = \left(- 300.4 \times {10}^{3}\right) + 8.31 \times 298 \times \ln \left(\frac{0.03}{0.001}\right)$

$\Delta {G}_{r} = \left(- 300.4 \times {10}^{3}\right) + 8422.66$

$\Delta {G}_{r} = - 291 , 977 \text{J}$

$\Delta {G}_{r} = - 292 \text{kJ}$

Just a side note about the reaction quotient $Q$. Here you can see that it is larger than ${K}_{c}$ (30 v 4.8). This means that there is a predominance of products so the equilibrium will shift to the left.

The reverse happens if $Q$ <${K}_{c}$