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In a catalyst experiment involving the Haber process `N_(2) + 3H_(2) rarr 2NH_(3)`, the rate of reaction was measured as Rate `= (Delta[NH_(3)])/(Delta t) = 2.0 xx 10^(-4) mol L^(-1) s^(-1)` What is the rate of reaction expressed in terms of (a) `N_(2)` (b) `H_(2)` ?A. `2.50 xx 10^(-4) mol L^(-1) s^(-1)`B. `1.25 xx 10^(-4) mol L^(-1) s^(-1)`C. `3.75 xx 10^(-4) mol L^(-1) s^(-1)`D. `5.00 xx 10^(-4) mol L^(-1) s^(-1)` |
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Answer» Correct Answer - C `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)` Rate `= (-d[N_(2)])/(dt) = (-d[H_(2)])/(3 dt) = (+d[NH_(3)])/(2dt)` `:. (-d[H_(2)])/(dt) = (3)/(2)(d[NH_(3)])/(dt)` `= (3)/(2) xx 2.5 xx 10^(-4) mol L^(-1) s^(-1)` `= 3.75 xx 10^(-4) mol L^(-1) s^(-1)`Correct Answer - C `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)` Rate `= (-d[N_(2)])/(dt) = (-d[H_(2)])/(3 dt) = (+d[NH_(3)])/(2dt)` `:. (-d[H_(2)])/(dt) = (3)/(2)(d[NH_(3)])/(dt)` `= (3)/(2) xx 2.5 xx 10^(-4) mol L^(-1) s^(-1)` `= 3.75 xx 10^(-4) mol L^(-1) s^(-1)` |
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