Question

Consider the reaction: N2(g) + 3H2(g) → 2NH3(g) Suppose that a particular moment during the reaction, molecular hydrogen is reacting at a rate of −0.0160 M/s. At what rate is ammonia being formed?

Answer 1

The ration of formation of ammonia gas is 0.01067 M/s.

Step-by-step explanation:

`N_2(g) + 3H_2(g)\rightarrow 2NH_3(g)`

Rate of the reaction = R

`R=(-1)/(1)(d[N_2])/(dt)=(-1)/(3)(d[H_2])/(dt)=(1)/(2)(d[NH_3])/(dt)`

Rate at which hydrogen gas is reacting =
`-(d[H_2])/(dt)=0.0160M/s`

Rate at which ammonia gas is forming =
`-(d[NH_3])/(dt)`

`(-1)/(3)(d[H_2])/(dt)=(1)/(2)(d[NH_3])/(dt)`

`(-1)/(3)(d[H_2])/(dt)* 2=(d[NH_3])/(dt)`

`(1)/(3)(-(d[H_2])/(dt))* 2=(d[NH_3])/(dt)`

`(2)/(3)* 0.0160M/s=0.01067 M/s`

The ration of formation of ammonia gas is 0.01067 M/s.