Question

Considering the following reaction: $CO(g) + 2{H_2}(g) \to C{H_3}OH(g)$ Suppose that the initial concentrations of the reactants are $[CO] = 0.500M$ and $[{H_2}] = 1.00M$ . Assuming that there is no product at the beginning of the reaction and that at equilibrium $[CO] = 0.15M$ , what is the equilibrium constant at this new temperature?

Answer 1

Hint: For solving this question, we have to first write the balanced chemical equation with an ICE table. ICE tables help us to find equilibrium concentrations of unknown compounds. After finding the value, we will substitute this value in the formula of equilibrium. Hence we get the final answer.
Formula used:
${K_C} = \dfrac{{{{[C]}^c} \times {{[D]}^d} \times ....}}{{{{[A]}^a} \times {{[B]}^b} \times .....}}$
$for\;a{A_{(aq)}} + b{B_{(aq)}} + .... \rightleftharpoons c{C_{(aq)}} + d{D_{(aq)}} + ....$
Where, ${K_c}$ = equilibrium constant
$A,B,....$ = reactants
$C,D,....$ = products
$[A]$ = equilibrium concentration of $A$ in moles
$a$ = number of moles of $A$

Complete answer:
First of all we have to write a balanced chemical equation with the ICE table. The ICE table which is known as the Initial, Change, Equilibrium table, is used to simplify the calculations in reversible equilibrium reactions, e.g. weak acids and weak bases or complex ion formation.
The balanced chemical equation is:
$CO(g) + 2{H_2}(g) \rightleftharpoons C{H_3}OH(g)$
And the ICE table is:

	$[CO]$	$[{H_2}]$	$[C{H_3}OH]$
Initial	$0.500$	$1.00$	$0$
Change	$ - x$	$ - 2x$	$ + x$
Equilibrium	$0.500 - x$	$0.100 - 2x$	$x$

So, at the equilibrium $[CO]$ = $0.15mol{L^{ - 1}}$
$ \Rightarrow 0.15mol{L^{ - 1}} = (0.500 - x)mol{L^{ - 1}}$
So, $x = 0.500 - 0.15$
$ \Rightarrow x = 0.35$
Then by substituting the value of $x$ in the $[{H_2}]$ , we will get:
$[{H_2}]$=$(0.100 - 2x)mol{L^{ - 1}}$
$ \Rightarrow (0.100 - 2 \times 0.35)mol{L^{ - 1}}$
$ = 0.30mol{L^{ - 1}}$
Now similar by substituting the value of $x$ in the $[C{H_3}OH]$ , we will get the value which is equal to:
$[C{H_3}OH]$= $x\;mol{L^{ - 1}}$
$ \Rightarrow 0.35mol{L^{ - 1}}$ .
After getting all the values, substitute them in the given formula of equilibrium constant, we get:
$\Rightarrow {K_c} = \dfrac{{[C{H_3}OH]}}{{[CO]{{[{H_2}]}^2}}}$
$ \Rightarrow \dfrac{{0.35}}{{0.15 \times {{(0.30)}^2}}}$
$ = 26$
So, the equilibrium constant at the new temperature is $26$ .

Note:
We should remember that the only thing which changes equilibrium is a change of temperature. It is independent of the actual quantities of reactants and products. The position of equilibrium is not changed by changing a catalyst.