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For the reaction, 2NO + H(2) rarr N(2)O ...

For the reaction, `2NO + H_(2) rarr N_(2)O + H_(2)O` , the rate of reaction was found to be `1.56 Pa s^(-1)` for a pressure of 373 Pa of NO and 0.25 Pa `s^(-1)` for a pressure of 152 Pa of NO. The pressure of `H_(2)` being constant. If pressure of NO was kept constant, the rate of reaction was found `1.60 Pa s^(-1)` for a pressure of `H_(2)` 289 Pa and `0.79 Pa s^(-1)` for a pressure of 144 Pa of `H_(2)`. Calculate the order of reaction.

Text Solution

Verified by Experts

The correct Answer is:
3
Rate of reaction `-(1)/(2)(dP_(NO))/(d t) = 1.5 = K(373)^(m)(P_(H_(2)))^(n)`
`-(1)/(2)(dP_(NO))/(d t) = 0.25 = K(152)^(m)(P_(H_(2)))^(n), (1.5)/(0.25) ((373)/(152))^(n) therefore m = 2`
For `P_(NO)` constant `-(dP_(H_(2)))/(d t) = 1.60 = (P_(NO))^(m) (289)^(n)`
`- (dP_(H_(2)))/(d t) = 0.79 (P_(NO))^(m) (144)^(n) therefore (1.60)/(0.79) = ((289)/(144))^(n) therefore n= 1`
Rate `= K[P_(NO)]^(2)[P_(H_(2))]^(1) therefore` Order of reaction `= 2 + 1 = 3`
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