Question

X2(g) + Y2(g) $\rightleftharpoons$ 2XY(g) reaction was studied at a certain temperature. In the beginning 1 mole of X2 was taken in a one litre flask and 2 moles of Y2 was taken in another 2 litre flask and both these containers are connected so equilibrium can be established. What is the equilibrium concentration of X2 and Y2? Given Equilibrium concentration of [XY] = 0.6 moles/litre.

• A. $\left( \frac { 1 } { 3 } - 0.3 \right) , \left( \frac { 2 } { 3 } - 0.3 \right)$
• B. $\left( \frac { 1 } { 3 } - 0.6 \right) , \left( \frac { 2 } { 3 } - 0.6 \right)$
• C. (1 – 0.3) , (2 – 0.3)
• D. (1 – 0.6) , (2 – 0.6)

Answer 1

Answer: (A)

$\left( \frac { 1 } { 3 } - 0.3 \right) , \left( \frac { 2 } { 3 } - 0.3 \right)$

Answer 2

X2 + Y2 $\rightleftharpoons$ 2XY

2x 2x = 0.6

$\Rightarrow$ x = 0.3

[x2] = $\frac { 1 } { 3 }$- 0.3 [y2] = $\frac { 2 } { 3 }$- 0.3

Therefore, (1) option is correct.