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InterviewSolution
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The decomposition of `NH_(3)` on platinum surface is zero order reaction the rate of production of `H_(2)` is `(k=2.5xx10^(-4) M s^(-1))`A. `3.35xx10^(-4) and 1.25xx10^(-4),` respectivelyB. `1.25xx10^(-4) and 3.75xx10^(-4),` respectivelyC. `3.75xx10^(-3) and 2.45xx10^(-3),` respectivelyD. `1.25xx10^(-3) and 3.25xx10^(-3),` respectively |
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Answer» Given,`k=2.5xx10^(-4) molL^(-1) S^(-1)` `2H_(3)(g)toN_(2)(g)+3H_(2)(g)` On diving the Eq.by 2, `NH_(3)(g)to(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)` Rate=`(d[NH_(3)])/(dt)=+(2d[N_(2)])/(dt)=(2)/(3)(d[H_(2)])/(dt)` For zero order reaction, rate of reaction =k so,`(-d[NH_(3)])/(dt)=(2d[N_(2)])/(dt)=(2)/(3)(d[H_(2)])/(dt)=2.5xx10^(-4)" ""mol"L^(-1)S^(-1)` `therefore "Rate of production of" N_(2)=(d[N_(2)])/(dt)=(k)/(2)` `=2.5xx10^(-4)" ""mol"L^(-1)S^(-1)` `therefore "Rate of production of" H_(2)=(d[H_(2)])/(dt)=(3)/(2)xx(2.5xx10^(-4)molL^(-1)S^(-1))` `=3.75xx10^(-4)molL^(-1)S^(-1)` |
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