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Question

The value of $$K_p$$ for dissociation of

    $$2HI{(g)}\Longleftrightarrow H_2{(g)} +I_2{(g)}$$

 is $$ 1.84\times 10^{-2}$$. If the equilibrium concentration of $$H_2$$ is 0.4789 mol litre$$^{-1}$$, calculate the concentration of $$HI$$ at equilibrium.


A
3.53 mol lit1
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B
0.53 mol lit1
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C
35.3 mol lit1
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D
0.353 mol lit1
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Solution

The correct option is A 3.53 mol lit$$^{-1}$$
As $$\Delta n=0$$, $$K_p=K_c=1.84 \times 10^{-2}$$. 

Also at equilibrium, $$[H_2][I_2]=0.4789M$$.

The expression for the equilibrium constant becomes $$K_c=\dfrac {[H_2][I_2]} {[HI]^2}$$.

Substitute values in the above expression.

$$1.84 \times 10^{-2}= \dfrac {0.4789 \times 0.4789} {[HI]^2}$$.

$$[HI]=3.53 \ mol\  lit^{-1}$$.

Chemistry

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