At `700 K` equilibrium constant for the reaction, `H_(2(g))+I_(2(g))hArr2HI_((g))`
is `54.8`. If `0.5 mol litre^(-1)` of `HI_((g))` is present at equilibrium at `700 K`, what are the concentrations of `H_(2(g))` and `I_(2(g))`, assuming that we initially started with `HI_((g))` and allowed it to reach equilibrium at `700 K`.

Explanation

To solve the problem step by step, we will follow the equilibrium expression and the information provided.

### Step 1: Write the equilibrium expression
For the reaction:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
...