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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.

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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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Knowledge Check

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

    A
    `approx 3`
    B
    slightly less than 7
    C
    slightly greater than 7
    D
    `=7`

  • For the equilibrium N_(2)+3H_(2)hArr2NH_(3),K_(c) at 1000K is 2.73xx10^(-3) if at equlibrium [N_(2)]=2M,[H_(2)]=3M, the concentraion of NH_(3) is

    A
    `0.00358` M
    B
    `0.0358` M
    C
    `0.358` M
    D
    `3.58` M

  • The equilibrium constant for the reaction H_(2)(g)+I_(2)(g)hArr 2HI(g) is 32 at a given temperature. The equilibrium concentration of I_(2) and HI are 0.5xx10^(-3) and 8xx10^(-3)M respectively. The equilibrium concentration of H_(2) is

    A
    `1xx10^(-3)M`
    B
    `0.5xx10^(-3)M`
    C
    `2xx10^(-3)M`
    D
    `4xx10^(-3)M`
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