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The following concentrations were obtain...

The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the equilibrium constant.

Text Solution

Verified by Experts

The correct Answer is:
`3.56xx10^2`
`K_c=([NH_3]^2)/([N_2][H_2]^3)`
`=(1.2xx10^(-2))^2/((1.5xx10^(-2))xx(3.0xx10^(-2))^3)`
`=3.56xx10^2`
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The following concentrations were obtained for the formation of NH_(3)" from " N_(2) and H_(2) at 500 K : [N_(2)]= 1.5 xx 10^(-2) M, [H_(2)] = 3.0 xx 10^(-2) M and [NH_(3)] = 1.2 xx 10^(-2) M. Calculate equilibrium constant.

The following concentration were obtained for the formation of NH_3 from N_2 and H_2 at equilibrium for the reaction N_2(g) +3H_2(g) hArr 2NH_3(g) [N_2]=1.5 xx 10^(-2) M [H_2]=3.0 xx 10^(-2) M [NH_3]=1.2 xx 10^(-2) M Calculate equilibrium constant.

Knowledge Check

  • The pH of 1 xx 10^(-3) M H_(2)O_(2) solution (K_(a)=2.2 xx 10^(-12)) is

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    `approx 3`
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