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Write the rate law. 5.The reaction below is first order with respect to bromide

ID: 549036 • Letter: W

Question

Write the rate law. 5.The reaction below is first order with respect to bromide ion, first order with respect to hydrogen ion and first order with respect to hydrogen peroxide. Rate Iral rareaction At3BC+D, was measured using different initial for the reaction? (b) What is the numerical value of the rate constant k? 6.The in the table below. (a) What is the concentrations of A and B. rate law Results are summarized in 0.100 5.40 10-4 0.100 0.200 o1004.32 10 0.200 0.200 4.32-10.3 Rate = 7.The initial rate of a reaction, A B+C2D E, was measured using different initial concentrations of A, B and C. Results are summarized in the table below. (a) What is the rate law for the reaction? (b) What is the numerical value of the rate constant k? (c) What will be the initial rate in an experiment using initial concentrations A 0.250 M, B = 0.230 M and C 0.1 25 M, BL 0.100 0.100 ICL Initial Rate. M/s 0.100 0.200 7.3 × 10-11 -9 2 0.2000.200 0.100 2.3 x 109 0.100 0.300 1,1 × 10 Rate- Init rate =

Explanation / Answer


5)

2 Br- + 2 H+ + H2O2 ------> Br2 + 2 H2O

Rate law = K[Br-]^1[H+]^1[H2O2]^1

6)

A + 3B -----> C + D

Rate = K[A]^x[B]^y

where x and Y are order with respect to A and B respectively

from table,

Rate1/Rate2 = K[A1]^x[B1]^y/K[A2]^x[B2]^y

5.4 * 10^-4/(4.32*10^-3) = (0.1)^x(0.1)^y/(0.2)^x*(0.1)^y

1/8 = (1/2)^x

x = 3 order with respect to A = 3

Rate2/Rate3 = K[A2]^x[B2]^y/K[A3]^x[B3]^y

(4.32*10^-3)/(4.32*10^-3) = (0.2)^x(0.1)^y/(0.2)^x*(0.2)^y

1 = (1/2)^y

y = 0 order with respect to B = 0

Rate = K[A]^3[B]^0

K = 5.4*10^-4/((0.1)^3(0.1)^0)

K = 0.54

7)

A + B + C -----> 2D + E

Rate = K [A]^x [B]^y [C]^z

Rate1/Rate2 = K [A1]^x [B1]^y [C1]^z/K [A2]^x [B2]^y [C2]^z

7.3*10^-11/(1.2*10^-9) = (0.1)^x*(0.1)^y*(0.1)^z/(0.2)^x*(0.1)^y*(0.1)^z

0.06083 = (1/2)^x

x = 4.04 order with respect to A

Rate2/Rate3 = K [A2]^x [B2]^y [C2]^z/K [A3]^x [B3]^y [C3]^z

(1.2*10^-9)/(2.3*10^-9) = (0.2)^x*(0.1)^y*(0.1)^z/(0.2)^x*(0.2)^y*(0.1)^z

0.5217 = (1/2)^y

y = 0.9387 order with respect to B

Rate2/Rate4 = K [A3]^x [B3]^y [C3]^z/K [A4]^x [B4]^y [C4]^z

(1.2*10^-9)/(1.1*10^-9) = (0.2)^x*(0.1)^y*(0.1)^z/(0.2)^x*(0.1)^y*(0.3)^z

1.091 = (1/3)^z

z = -0.0793

Rate = K[A]^4.04 *[B]^0.9387 * [C]^(-0.0793)

K = (0.1)^4.04 *(0.1)^0.9387 * (0.1)^(-0.0793)/(7.3*10^-11)

K = 1.7269 * 10^5

Rate = 1.7269 * 10^5 (0.25)^4.04 * (0.23)^0.9387*(0.125)^(-0.0793)

Rate = 189.414