Question
For the reaction 2 NOBr _ (g) 2 NO _ (g) + Br _ 2(g) rate law is r = k[ NOBr ]^2, if rate constant is 1 62 mol dm ^ -3 s ^ -1 and concentration of NOBr is 2 10^ -3 mol L ^ -1 . What is the rate of reaction ?
For the reaction 2 NOBr _ (g) 2 NO _ (g) + Br _ 2(g) rate law is r = k[ NOBr ]^2, if rate constant is 1 62 mol dm ^ -3 s ^ -1 and concentration of NOBr is 2 10^ -3 mol L ^ -1 . What is the rate of reaction ?
B. 6 48 10^ -6 mol dm ^ -3 s ^ -1
Given rate law is r = k[ NOBr ]^2 Substituting the given values: k = 1.62 [ NOBr ] = 2 10^ -3 mol L ^ -1 = 2 10^ -3 mol dm ^ -3 r = 1.62 (2 10^ -3 )^2 r = 1.62 4 10^ -6 r = 6.48 10^ -6 mol dm ^ -3 s ^ -1 Answer: 6 48 10^ -6 mol dm ^ -3 s ^ -1
Related: Chemistry — Chemical Kinetics · All PYQ Banks