Question

Ammonia and oxygen react at high temperature as:

4NH_3(g) + 5O_2(g) → 4NO_(g) + 6H₂O_(g) In an experiment, rate of formation of NO_(g) is 3.6 x 10^-3 mol L^-1 s ^-1.

Calculate:

a. Rate of disappearance of ammonia.

b. Rate of formation of water.

Answer 1

Rate of formation of NO(g) = d[NO]/dt= 3.6 x 10^–3 mol L^–1 s^–1

Rate of reaction `=1/4 (d[NH_3])/dt=-1/5 (d[O_2])/dt=+1/4(d[NO])/dt=+1/6(d[H_2O])/dt`

`1/4(d[NO])/dt=1/4xx3.6xx10^(-3)=9.0xx10^(-4)mol L^(-1)s^(-1)`

a. Rate of disappearance of `NH_3=-(d[NH_3])/dt`

`-1/4(d[NH_3])/dt=1/4(d[NO])/dt=9.0xx10^(-4) mol L^(-1)s^(-1)`

`-(d[NH_3])/dt=3.6xx10^(-3)mol L^(-1)s^(-1)`

Rate of formation of `H_2O =(d[H_2O])/dt`

`1/6(d[H_2O])/dt=1/4(d[NO])/dt=9.0xx10^(-4) mol L^(-1)s^(-1`

 

`(d[H_2O])/dt=5.4xx10^(-3)mol L^(-1)s^(-1)`