For the following reaction,
`2H_2 +2NO hArr 2H_2 O +N_2 `
the following rate data was obtained

Determine the rate equation and calculate the value of rate constant, k.

Answer

Rate ` =k [NO ]^2 [H_2] ; k = 0.075 mol^(-2) L^2 s^(-1)`

Explanation

Rate of reaction is given as: Rate` = k [NO]^x [H_2]^y`
(i) For experiment `1, 4.8 xx 10^(-3) = k [0.4]^x (0.4)^y`
(ii) For experiment `2, 19.2 xx 10^(-3) =k(0.80)^x (0.4)^y`
(iii) For experiment `3,9.6 xx 10^(-2) =k (0.40)^x (0.80)^y`
Divding eq. (i) by (ii), we get
`( 4.8 xx 10^(-3))/( 19.2 x 10^(-3))= (k (0.4 )^x ( 0.4 )^y)/(19.2 xx 10^(-3) )=( k ( 0.80 )^x ( 0.4 )^y)`
`or 1/4 = (1/2)^x or (1/2)^2 = (1/2)^x`
` therefore x=2`
Dividing eq. (i) by (iii), we get
`( 4.8 xx 10^(-3))/( 9.6 xx 10^(-3)) = (k ( 0.4 ) ^x (0.4 )^y)/( k(0.4 )^x ( 0.80)^y) or (1/2)^1 = (1/2)^y`
Rate equation : `k [NO]^2 [H_2]`
From experiment `1, 4.8 xx 10^(-3) =k (0.4)^2 (0.4)^1`
`or k=( 4.8 xx 10^(-3) )/((0.4)^3) = 0.075 mol^(-2) L^2 s^(-1)`