The answer is #0.53M#.
Starting from the balanced chemical equation
#H_(2(g)) + I_(2(g)) rightleftharpoons2HI_((g))#
Since the reaction's equlibrium constant, #K_(eq)#, greater than 1, the reaction will favor the formation of the product, #HI#, so we would expct the equilibrium concentrations 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: http://en.wikipedia.org/wiki/RICE_chart)
...#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...
