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The reaction : 2N_(2)O_(5)(g) to 2NO_(2)(g) + (O_(2))_(g) was studied and the following data were collected. Determine: (i) the order, (ii) the rate law and (iii) rate constant for the reaction. |
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Answer» Solution :Let the order of the REACTION = n `Rate propto [N_(2)O_(5)]^(n)` `(34 xx 10^(-5)) propto (1.13 xx 0^(-2))^(n)` `(25 xx 10^(-5)) propto (0.84 xx 10^(-2))^(n)` DIVIDING eqn. (i) by eqn. (ii), we have `(34 xx 10^(-5))/(25 xx 10^(-5)) = (1.13 xx 10^(-2))^(n)/(0.84 xx 10^(-2))^(n), (1.36)^(1) = (1.35)^(n)` or n=1 i) The order of the reaction = 1, ii) Rate law, The rate of reaction depends UPON one concentration team of `N_(2)O_(5)` because of the reaction is 1. ii) Rate law. The rate of reaction depends upon one concentration term of `N_(2)O_(5)` because order of the reaction is 1. Reaction Rate = `k[N_(2)O_(5)]` III) Rate constant, `k = "Reaction rate"/([N_(2)O_(5)])= (18 xx 10^(-5) mol L^(-1)min^(-1))/(0.62 xx 10^(-2)mol L^(-1))= 0.0.29 min^(-1)` |
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