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For the reaction, 2 NOBr(g) to 2NO2(g...

For the reaction, `2 NOBr(g) to 2NO_2(g) + Br_2(g)` ,
the rate law is rate = `k[NOBr]^2` . If the rate of the reaction is `6.5 xx 10^(-6) "mol L"^(-1) s^(-1)` when the concentration of NOBr is `2 xx 10^(-3) `mol `L^(-1)` . What would be the rate constant for the reaction?

Text Solution

Verified by Experts

rate `k[NOBr]^2 or k = ("rate")/([NOBr]^2)`
` = (6.5 xx 10^(-6) "mol" L^(-1) s^(-1))/((2 xx 10^(-3) "mol" L^(-1))^2)`
` = 1.625 "mol"^(-1) L^1 s^(-1)`
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