For the thermal decomposition of azomethane, `CH_3 N_2 CH_3` at 600 K to `N_2` and `C_2 H_6` the following data was obtained:

where `P_A` is the partial pressure of azomethane. Show that the decomposition is a first order reaction and find the rate constant.

Answer

`k= 3.6 xx 10^(-4) s^(-1)`

Explanation

Azomethane decomposes mainly according to the reaction
` CH_3 NNCH_3(g) to C_2 H_6 (g) +N_2`
If the reaction obeys first order kinetics it must obey the equation ,
`k=( 2.303 )/(t) log""(a)/( (a-x))`
` a prop P_0 and (a-x) prop P_A , P_0 = 8.20 xx 10^(-2) torr `
` or K = ( 2.303 )/(t) log"" (P_0)/(P_A)`
` (##KAL_KLC_ISC_CHE_XII_C04_SLV_085_S01.png" width="80%">
` k_1 = ( 2.303 )/( 1000 ) log"" (8.20 )/(5.72 )`
` = 2.303 xx 10^(-3) [ log 8.20 - log 5.72 ] `
` = 2.303 xx 10^(-3) [ 0.9138 - 0.7573]`
` = 2.303 xx 10^(-3) xx 0.1565 sec^(-1) = 3.6 xx 10^(-4) sec^(-1)`
` k_2 = ( 2.303 )/( 2000 ) log"" ( 8.20 )/(3.99 )`
` = 1.1515 xx 10^(-3) xx [log 8.20 - log 3.99 ]`
` = 1.1515 xx 10^(-3) [ 0.9138 - 0.6099 ]`
`= 1.1515 xx 10^(-3) xx 0.3129 `
`3.60 xx 10^(-4) s^(-)`
` k_3 = ( 2.303 )/( 3000 ) log"" ( 8.20 )/( 2.78 ) = 7.6766 [ log 8.20 - log 2.78]`
`= 7.6766 xx 10^(-4) [ 0.9138 -0.4440 ]`
` = 7.6766 xx 0.49698 xx 10^(-4) s^(-1) = 3.6 xx 10^(-4) s^(-1)`
since first order the equation j gives a constant value of k reaction follows first order kinetics and k = `3.6 xx 10^(-4) s^(-1)`