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Express the rate of the following reaction in terms of different reactants and products : 4" NH"_(3)(g)+5" O"_(2)(g)to4" NO"(g)+6 H_(2)O(g) If the rate of formation of NO is 3*6xx10^(-3)" mol "L^(-1)s^(-1), calculate (i) the rate of disappearance of NH_(3) (ii) rate of formation of H_(2)O. |
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Answer» Solution :Rate of REACTION `=(1)/(4)(d[NH_(3)])/(dt)=-(1)/(5)(d[NO_(2)])/(dt)=+(1)/(4)(d[NO])/(dt)=+(1)/(6)([H_(2)O])/(dt)` Rate of disappearance of `NH_(3)`= Rate of FORMATION of `NO=3*6xx10^(-3)" MOL "L^(-1)x^(-1)` Rate of formation of `H_(2)O=(6)/(4)xx` Rate of formation of NO i.e., `""(d[H_(2)O])/(dt)=(3)/(2)xx(3.6xx10^(-3)"mol "L^(-1)s^(-1))=5.4xx10^(-3)"mol "L^(-1)s^(-1)` Alternatively, `(1)/(6)(d[H_(2)O])/(dt)=(1)/(4)(d[NO])/(dt)implies" Rate of formation of "H_(2)O,(d[H_(2)O])/(dt)=(6)/(4)xx(d[NO])/(dt)` |
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