InterviewSolution
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InterviewSolution
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when ammonia is treated with `O_(2)` at elevated temperature the rate of disappearance of ammonia is found to be `3.5 xx10^(-2)` mol `L^(-1) sec (-1)` during a measured time interval. The rate of appearance of water will beA. `3.5xx10^(-2)mol L^(-1)sec^(-1)`B. `5.3xx10^(-2) mol L ^(-1) s ^(-1)`C. `2.1 xx10^(1) mol L^(-1) sec ^(-1) `D. `1.4 xx 10^(-1) mol L^(-1) s^(-1)` |
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Answer» Correct Answer - B Since the relative rates of appearance of the products are governed by the coefficients of the chemical equation, the first step is to write the equation for the reaction: `4NH_(3)(g)+5O_(2)(g)rarr4NO(g)+6H_(2)O(g)` Notice that if `4mol` of `NH_(3)` disappears, then `6mol` of `H_(2)O` from or when `1mol` of `NH_(3)` disappease, then `6//4` or `1.5mol` of `H_(2)O` from. Thus, the rate of appearance of `H_(2)O` is `1.5` times as fast as the rate of disappearance of `NH_(3)` : `(Delta[H_(2)O])/(Deltat)=1.5((Delta[NH_(3)])/(Deltat))` `=(1.5)(3.5xx10^(-2molL^(-1)s^(-1))` `=5.25xx10^(-2)molL^(-1)s^(-1)` `=5.3xx10^(-2)molL^(-1)s^(-1)` |
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