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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
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B
0.818
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C
1.818
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D
1.182
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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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