InterviewSolution
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InterviewSolution
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For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.A. `1.25xx10^(-2)mol L^(-1)s^(-1) and 6.25xx10^(-3)mol^(-1)L^(-1)s^(-1)`B. `6.25xx10^(-3)mol L^(-1)s^(-1) and 6.25xx10^(-3)mol^(-1)L^(-1)s^(-1)`C. `1.25xx10^(-2)mol L^(-1)s^(-1)and 3.125xx10^(-3) mol^(-1)L^(-1)s^(-1)`D. `6.25xx10^(-3)mol L^(-1)s^(-1) and 3.125xx10^(-3)mol^(-1)L^(-1)s^(-1)` |
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Answer» Correct Answer - C For the given reaction `N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g)` the rate can be expressed in terms of any of the three species (reactant or products): Rate `=(-d[N_(2)O_(5)])/(dt)=(d[NO_(2)])/(2dt)=(d[O_(2)])/(1//2dt)` Thus, rate of information of `NO_(2)` is given as: `(d[NO_(2)])/(dt)=2((-d[N_(2))_(5)])/(dt))` `=2.(6.25xx10^(-3)molL^(-1)s^(-1))` `=12.5xx10^(-3)molL^(-1)s^(-1)` `=1.25xx10^(-2)molL^(-1)s^(-1)` The rate of formation of `O_(2)` is given as `(d[O_(2)])/(dt)=(1)/(2)(-(d[N_(2)O_(5)])/(dt))` `=(1)/(2)(6.25xx10^(-3)molL^(-1)s^(-1))` `=3.125xx10^(-3)molL^(-1)s^(-1)` |
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