JEE MAIN - Chemistry (2021 - 26th August Evening Shift - No. 14)
The equilibrium constant Kc at 298 K for the reaction A + B $$\rightleftharpoons$$ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ___________ $$\times$$ 10$$-$$2 M. (Nearest integer)
Answer
182
Explanation
_26th_August_Evening_Shift_en_14_1.png)
$$\therefore$$ $${K_C} = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$$
$$100 = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$$
$${{1 + x} \over {1 - x}} = 10$$
$$x = {9 \over {11}}$$
Moles of D = 1 + x
$$ = 1 + {9 \over {11}} = {{20} \over {11}}$$
$$ = 1.818 = 181.8 \times {10^{ - 2}} = 181.8 \times {10^{ - 2}}$$
$$ \cong 182 \times {10^{ - 2}}$$ M
Comments (0)
