1.

The decomposition of NH_3 on platinum surface is zero· order with rate constant k = 2.5 xx 10^(-4) mol L^(-1) s^(-1). The rate of production of H_2 will be

Answer»

`2.5 xx 10^(-4) mol L^(-1) s^(-1)`
`0.83 xx 10^(-4) mol L^(-1) s^(-1)`
`7.5 xx 10^(-4) mol L^(-1) s^(-1)`
`5.0 xx 10^(-4) mol L^(-1) s^(-1)`

Solution :`2NH_3 overset(Pt)to N_2+3H_2`
`"RATE"= 1/2(d[NH_3])/(dt)=(d[N_2])/(dt)= 1/3(d[H_2])/(dt)`
`=1/3 (d[H_2])/(dt)=k = 2.5 xx 10^(-4)mol L^(-1)s^(-1)`
Rate of production of `H_2`.
`:. (d[H_2])/(dt) = 3 xx 2.5 xx 10^(-4)`
`=7.5 xx 10^(-4)mol L^(-1)s^(-1)`


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