The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) mol litre^(-1)s^(-1)`?

The rate for the decomposition of `NH_(3)` on platinum surface is zero

1.	The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) mol litre^(-1)s^(-1)`?
Answer» `2NH_(3) rarr N_(2)+3H_(2)` `-1/2(d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=1/3(d[H_(2)])/(dt)` Rate `=K [NH_(3)]^(0)` or `(d[NH_(3)])/(dt)=2.5xx10^(-4) mol litre^(-1) s^(-1)` `(d[N_(2)])/(dt)=1/2xx2.5xx10^(-4)` `=1.25xx10^(-4) mol litre^(-1) s^(-1)` `(d[H_(2)])/(dt)=3/2xx2.5xx10^(-4)` `=3.75xx10^(-4) mol litre^(-1) s^(-1)`

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The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) mol litre^(-1)s^(-1)`?

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