Chemical EquilibriumHard
Question
For the reaction X2(g) + Y2(g) $\rightleftharpoons$2XY(g), 2 moles of ‘X2’ was taken in a 2 L vessel and 3 moles of ‘Y2’ was taken in a 3 L vessel. Both vessels were then connected. At equilibrium, the concentration of ‘XY’ is 0.7 M. The equilibrium concentrations of ‘X2’ and ‘Y2’ would be
Options
A.0.65 M, 0.65 M
B.0.30 M, 0.30 M
C.0.25 M, 1.25 M
D.0.05 M, 0.25 M
Solution
X2(g) + Y2(g) $\rightleftharpoons$2XY(g)
Initial moles 2 3 0
Final moles 2 – x 3 – x 2x
$\lbrack XY\rbrack = \frac{2x}{5} = 0.7 \Rightarrow x = 1.75 $$${\therefore\left\lbrack X_{2} \right\rbrack = \frac{2 - x}{5} = 0.05\text{ M} }{\text{and }\left\lbrack Y_{2} \right\rbrack = \frac{3 - x}{5} = 0.25\text{ M}}$$
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