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
B
0.53 mol lit1
C
35.3 mol lit1
D
0.353 mol lit1

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More