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Consider the reaction: `N_(2(g))+3H_(2(g))rarr 2NH_(3(g))`. The equally relationship between `-(d[NH_(3)])/(dt)` and `-(d[H_(2)])/(dt)` is:A. `(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`B. `+(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)`C. `+(d[NH_(3)])/(dt)=-(3)/(2)(d[H_(2)])/(dt)`D. `(d[NH_(3)])/(dt)=-(d[H_(2)])/(dt)` |
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Answer» Correct Answer - B for the reactions , `N_(2)(g) +3H_(2)(g) to 2NH_(2)(g)` the rate of reactionn wrt `N_(2)=-(d[N_(2)])/(dt)`[ Rate of disappearance ] the rate of reaction with respect to `H_(2)=-(1)/(3)(d[H_(2)])/(dt)`[rate of disappearance ] the rate of reaction with respect to `NH_(3)=+(1)/(2)(d[NH_(3)])/(dt)`[rate of appearance ] hence at a fixed time `-(d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=+(1)/(2)(d[NH_(3)])/(dt)` `or +(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)` `or +(d[NH_(3)])/(dt)=-(2d[N_(2)])/(dt)` |
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