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Calculate the percent dissociation of H(...

Calculate the percent dissociation of `H_(2)S(g)` if `0.1 mol` of `H_(2)S` is kept in `0.4 L` vessel at `1000 K`. For the reaction:
`2H_(2)S(g) hArr 2H_(2)(g)+S_2(g)`
The value of `K_(c )` is `1.0xx10^(-6)`

A

0.05

B

0.02

C

0.06

D

0.08

Text Solution

Verified by Experts

The correct Answer is:
B
`2H_(2)S(g) rarr 2H_(2)(g) + S_(2)(g)`
`k_(c) = [H_(2)]^(2)[S_(2)]/[H_(2)S]^(2) = [x/V]^(2)[x/2V]/[[0.1-x]/v^(2)]`
`k_(c) = x^(3)/(2V(0.1)^(2)) = 1.0 xx 10^(-6)`
`x^(3) = 1.0 xx 10^(-6) xx 2 xx 0.4 xx 0.1 xx 0.1 = 8 xx 10^(-9)`
`x = 2 xx 10^(-3)`
`% dissociation = 2 xx 10^(-3) xx 100/0.1 = 2%`
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