Question #74a96

Jul 8, 2017

$2.40 \cdot {10}^{33}$

Explanation:

You know that at ${25}^{\circ} \text{C}$, the reaction

$2 {\text{HCl"_ ((g)) rightleftharpoons "H"_ (2(g)) + "Cl}}_{2 \left(g\right)}$

has an equilibrium constant equal to

${K}_{c} = 4.17 \cdot {10}^{- 34}$

The very, very small value of ${K}_{c}$ for the forward reaction tells you that at this temperature, the equilibrium will lie far to the left, meaning that almost all the hydrogen chloride present in the sample will not decompose to produce hydrogen gas and chlorine gas.

This implies that at this temperature, the reverse reaction will be highly favored, meaning that if you start with hydrogen gas and chlorine gas, you should expect the reaction vessel to contain almost exclusively hydrogen chloride.

So even without doing any calculation, you should be able to look at this equilibrium and say that the equilibrium constant for the reverse reaction, ${K}_{c}^{'}$, is significantly greater than $1$.

${K}_{c}^{'} \text{ >> } 1 \to$ the $\text{>>}$ symbol just means significantly greater than

Now, you know that the equilibrium constant for the forward reaction is equal to

${K}_{c} = \left({\left[\text{H"_2] * ["Cl"_2])/(["HCl}\right]}^{2}\right)$

The reverse reaction looks like this

${\text{H"_ (2(g)) + "Cl"_ (2(g)) rightleftharpoons 2"HCl}}_{\left(g\right)}$

This time, the equilibrium constant is equal to

${K}_{c}^{'} = \left(\left[{\text{HCl"]^2)/(["H"_2] * ["Cl}}_{2}\right]\right)$

Notice that the equilibrium constant for the reverse reaction is simply the reciprocal of the equilibrium constant for the forward reaction, since

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{\left[{\text{H"_2]))) * color(red)(cancel(color(black)(["Cl"_2]))))/color(red)(cancel(color(black)(["HCl"]^2))) * color(red)(cancel(color(black)(["HCl"]^2)))/(color(red)(cancel(color(black)(["H"_2]))) * color(red)(cancel(color(black)(["Cl}}_{2}\right]}}}\right) = 1$

So, you know that

${K}_{c} \cdot {K}_{c}^{'} = 1$

or

${K}_{c}^{'} = \frac{1}{K} _ c$

Plug in your value to get

${K}_{c}^{'} = \frac{1}{4.17 \cdot {10}^{- 34}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.40 \cdot {10}^{33}}}}$

The answer is rounded to three sig figs.

As predicted, you do have

${K}_{c}^{'} \text{ >> } 1$