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when ammonia is treated with O(2) at ele...

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 be

A

`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)`

Text Solution

Verified by Experts

The correct Answer is:
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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