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Ammonia and oxygen react at high temperature as: `4NH_(3) (g) + 5O_(2)(g) rarr 4NO(g) + 6H_(2)O (g)` In an experiment, the rate of formation of `NO` is `3.6 xx 10^(-3) Ms^(-1)`. Calculate (a) the rate of disappearance of ammonia and (b) the rate of formation of water. |
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Answer» For the reaction `4NH_(3)(g)+5O_(2) (g) rarr 4NO(g)+6H_(2)O(g)` Rate `=-1/4 (d[NH_(3)])/(dt)` `=1/5 (d[O_(2)])/(dt)=1/4 (d[NO])/(dt)` `=1/6 (d[H_(2)O])/(dt)` (i) Rate of disappearance of ammonia Rate of disappearance of `NH_(3)` = Rate of formation of NO `=3.6xx10^(-3)` mol `L^(-1) S^(-1)` (ii) Rate of formation of water Rate of formation of `H_(2)O=(d[H_(2)O])/(dt)` From rate, equation, `1/6 (d[H_(2)O)])/(dt)=1/4 (d[NO])/(dt)` `:. (d[H_(2)O])/(dt)=6/4 xx3.6 xx10^(-3)` `=5.4xx10^(-3)` mol `L^(-1) S^(-1)` |
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