# How do you find the equilibrium constant for the reverse reaction of this problem?

## Given : Temperature : 1500 degrees Celcius Equilibrium Constant : 0.1764 Balanced equation: $C O + 3 {H}_{2} = C {H}_{4} + {H}_{2} O$ Need: Equilibrium Constant for $C {H}_{4} + {H}_{2} O = C O + 3 {H}_{2}$

Aug 19, 2016

$K ' = 5.670$

#### Explanation:

The equilibrium constant for the reaction:

$C O + 3 {H}_{2} r i g h t \le f t h a r p \infty n s C {H}_{4} + {H}_{2} O$

is written as:

$K = \frac{\left[C {H}_{4}\right] \left[{H}_{2} O\right]}{\left[C O\right] {\left[{H}_{2}\right]}^{3}} = 0.1764$

The equilibrium constant for the opposite reaction:

$C {H}_{4} + {H}_{2} O r i g h t \le f t h a r p \infty n s C O + 3 {H}_{2}$

is written as:

$K ' = \frac{\left[C O\right] {\left[{H}_{2}\right]}^{3}}{\left[C {H}_{4}\right] \left[{H}_{2} O\right]} = \frac{1}{K} = \frac{1}{0.1764} = 5.670$

Here is a video that explains this topic in details:
Chemical Equilibrium | The Equilibrium Conditions.