The answer is $0.53M$.
Starting from the balanced chemical equation
$H_(2(g)) + I_(2(g)) rightleftharpoons2HI_((g))$
Since the reaction's , $K_(eq)$, greater than 1, the reaction will favor the formation of the product, $HI$, so we would expct the equilibrium of $H_2$ and $I_2$ to be smaller than the concentration of $HI$.
From the data given we can determine the starting concentrations of both $H_2$ and $I_2$ to be
$C_(H_2) = n_(H_2)/V = (2.40 mol es)/(1.00L) = 2.40M$
$C_(I_2) = n_(I_2)/V = (2.40 mol es)/(1.00L) = 2.40M$
We can now determine the equilibrium concentrations for this reaction by using the ICE method (more here: )
...$H_(2(g)) + I_(2(g))rightleftharpoons2HI_((g))$
I: 2.40........2.40...........0
C: (-x).............(-x)..........(+2x)
E: 2.40-x.....2.40-x........2x
We know that $K_(eq) = ([HI]^2)/([H_2]*[I_2])$, so we get
$K_(eq) = (2x)^2/((2.40-x)(2.40-x)) = (4x^2)/(2.40-x)^2 = 49.0$
Rearranging this equation will give us
$45x^2 - 235.2x + 282.24 = 0$, which produces two values for $x$, $x_1 = 3.36$ and $x_2 = 1.87$; we cannot choose $x_1$, since that would imply negative concentration values at equilibrium for both $H_2$ and $I_2$ (2.40 -3.36 = -0.96);
Therefore, $x = 1.87$, which means that, at equilibrium,
$[H_2] = 2.40 - 1.87 = 0.53M$
$[I_2] = 2.40 - 1.87 = 0.53M$
${HI] = 2 * 1.87 = 3.74M$
Notice how the final concentrations match the estimate derived from $K_(eq)$'s value - the reaction indeed favors the product ,$HI$.
A quick word on the ICE table...the starting concentration of $HI$ is 0 M because only $H_2$ and $I_2$ are present in the vessel; x simply represents the change in concentrations in accordance to the balanced chemical equation -> 1 mole of $H_2$ and 1 mole of $I_2$ combine to form 2 moles of $HI$ - that is where the +2x comes from...
