Question

Exercise 14.40 - Enhanced - with Feedback 10 of 32 > A Review | Constants Periodic Table The following reaction is first order in N, 0, N20(g) NO: (C) + NO2(g) The rate constant for the reaction at a certain temperature is 0.053/s. What would the rate of the reaction be at the same concentration as in Part A if the reaction were second order? (Assume the same numerical value for the rate constant with the appropriate units.) Express your answer using two significant figures. You may want to reference (Pages 593 - 598) Section 14.4 while completing this problem. ALD O O ? Tate = rate= M /S Submit Request Answer Part C What would the rate of the reaction be at the same concentration as in Part A if the reaction were zero order? Express your answer using two significant figures. Y ALP R O 21 ? rate = Mis Submit Request Answer
Write a rate law for the reaction. Write a rate law for the reaction. Rate=k Rate=k[A] Rate=k[A]2 Rate=k[A]3

Answer 1

it is given to assume rate constant k to be same in all case.

numerical value is same but it unit is different for different order.

k= rate/[A]^n

=(Ms-1)/(M^n)

you will get the unit for different order.

n represent order.

M represent Molarity.

s represent time.

Soln k = 0.033s- knit of k is set thing it is 1st order. Part A Rate = K[AL Rate = 0.053 (5.7x10-2) ---300 81031 Given K = 0.053 same as Part A. - so K = 0.053 M-tgat (sos and ordes) Rate = K[A] = 0.053 M-15-tx/ s.7x10-2) ? M2 - 1.7 X10-4 M Sat for zero osder. - K = 0.053 mst Rate = K[A] Rate =k. Rate = 10.053 / Mst