A reaction has the rate law Rate = k[A][B]^2. (a) What is the order of the react

Question

A reaction has the rate law Rate = k[A][B]^2. (a) What is the order of the reaction with respect to A? (b) What is the order of the reaction with respect to B? (c) What is the overall order of the reaction? (d) What happens to the rate if [A] is halved? (e) What happens to the rate if [B] is tripled? The following data were collected for the reaction: H_2(g) + 2NO(g) rightarrow N_2O(g) + H_2O(g) (a) Determine the rate law. (b) Determine the rate constant (with units). (c) Determine the rate if [NO] = 0.50 M and [H_2] = 0.50 M. (d) By what factor does the initial rate change if [NO] is changed from 0.015 M to 0.040 M?

Explanation / Answer

Rate law for the reaction is given by = K[A] [B]2

The powers of reactants gives the order with respect to the reactants

Hence the reaction order is 1 with respect to A, 2 with respect to B and overall order is sum of the powers of concentration terms , hence 1+2=3

When A is halved with out changing [B], the new rate= K([A]/2)*[B]2 = original rate/2

2.

From experiment -1, K[0.35]a[0.3]b = 2.835*10-3   (1)

From experiment-2 , K[0.35]a[0.60]b = 1.134*10-2   (2)

Eq. 2/Eq.1 gives 2b= 4, b= 2

From experiment-3, K[0.7]a[0.60]b = 2.268*10-2   (3)

Eq.3/Eq.2 gives [2]a = 2, a=1

So the rate law becomes r= K[H2][NO]2

From Eq.1,

K[0.35] [0.3]2 = 2.885*10-3 , K=0.092 /M2.sec

The rate law becomes r= 0.092 [H2] [NO]2

At[ H2] =[NO] =0.5M

Rate = 0.092[0.5] [0.5]2 = 0.0115 M/sec

Without changing [H2], if [NO] is changed from 0.015M to 0.040M

Rate1 = K[H2] [0.015]2 for 0.015M of NO and

Rate 2= K[ H2] [0.040]2 for 0.040M NO

Rate2/rate 1= 7.11

Rate = 7.11 times rate 1

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