JEE MAIN - Chemistry (2016 (Offline) - No. 8)
The equilibrium constant at 298 K for a reaction A + B $$\leftrightharpoons$$ C + D is 100. If the initial concentration of
all the four species were 1M each, then equilibrium concentration of D (in mol L–1) will be:
0.818
1.818
1.182
0.182
Explanation
Given,
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$$\therefore$$ $$\,\,\,{K_c} = {\left( {{{1 + a} \over {1 - a}}} \right)^2} = 100$$
$$\therefore$$ $$\,\,\,{{1 + a} \over {1 - a}} = 10$$
On solving
$$a=0.81$$
$${\left[ D \right]_{At\,eq}} = 1 + a = 1 + 0.81 = 1.81$$
$$\therefore$$ $$\,\,\,{K_c} = {\left( {{{1 + a} \over {1 - a}}} \right)^2} = 100$$
$$\therefore$$ $$\,\,\,{{1 + a} \over {1 - a}} = 10$$
On solving
$$a=0.81$$
$${\left[ D \right]_{At\,eq}} = 1 + a = 1 + 0.81 = 1.81$$
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