Question

# The equilibrium constant at $$298K$$ for a reaction $$A+B\rightleftharpoons C+D$$ 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:

A
0.182
B
0.818
C
1.818
D
1.182

Solution

## The correct option is C $$1.818$$Solution:$$A\quad +\quad B\quad \rightleftharpoons \quad C\quad +\quad D$$$$1\quad \quad \quad 1\quad \quad \quad 1\quad \quad \quad 1$$$$1-x\quad 1-x\quad 1+x\quad 1+x\\$$$${ K }_{ c }=\frac { [C][D] }{ [A]\quad [B]\\ }$$$$=\frac { (1+x){ }^{ 2 } }{ (1-x){ }^{ 2 } } \\$$$$=100$$$$\Rightarrow \frac { 1+x }{ 1-x } =10$$$$\Rightarrow 1+x\quad =10-10x$$$$\Rightarrow 11x=9$$$$x=\frac { 9 }{ 11\\ \\ }$$$$[D]=1+x=1+\frac { 9 }{ 11\\ \\ }$$$$=1.818\ M$$ is the answer.Chemistry

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