The rate of the reaction, `N_(2(g)) + 3H_(2(g)) rarr 2NH_(3(g))`, was measured as
`(d)/(dt)[NH_3]= 2 xx 10^(-4)" mol "L^(-1)s^(-1)`
The rate of the reaction expresses in terms of `N_2` and `H_2` are

Options

A. `{:("Rates in terms of "N_2,"Rate in terms of "H_2),("mol "L^(-1)s^(-1),"mol "L^(-1)s^(-1)),(1xx10^(-4),3xx10^(-4)):}`
B. `{:("Rates in terms of "N_2,"Rate in terms of "H_2),("mol "L^(-1)s^(-1),"mol "L^(-1)s^(-1)),(3xx10^(-4),1xx10^(-4)):}`
C. `{:("Rates in terms of "N_2,"Rate in terms of "H_2),("mol "L^(-1)s^(-1),"mol "L^(-1)s^(-1)),(1xx10^(-4),1xx10^(-4)):}`
D. `{:("Rates in terms of "N_2,"Rate in terms of "H_2),("mol "L^(-1)s^(-1),"mol "L^(-1)s^(-1)),(2xx10^(-4),2xx10^(-4)):}`

Answer

A

Explanation

`N_(2(g)) + 3H_(2(g)) rarr 2NH_(3(g))`
Rate=-`(d[N_2])/(dt)=-1/3(d[H_2])/(dt)=1/2(d[NH_3])/(dt)`
`1/2(d[NH_3])/(dt)=-(d[H_2])/(dt)=(2xx10^(-4))/(2)=1xx10^(-4)`
`-(d[H_2])/(dt)=3/2(d[NH_2])/(dt)=3/2xx2xx10^(-4)=3xx10^(-4)`