(a) Write the rate law for the reaction 2A + B → C if the reaction (1) is second order in B and overall third order, –rA = (2) is zero order in A and first order in B, –rA = (3) is zero order in both A and B, –rA = (4) is first order in A and overall zero order. –rA = (b) Find and write the rate laws for the following reactions (1) H2 + Br2 → 2HBr (2) H2 + I2 → 2HI

A. Write the rate law for this reaction

2A + B > C

Rate law is expressed as

-rA = kaCa^αCb^β

n = α + β

Where, ka is the rate constant;

Ca is the concentration of reactant A, Cb is the conversation of reactant B, α is the rate order for reactant A, β is the rate order for reactant B, n is the overall order.

1. β = 2, n = 3; α = 3-2 = 1

Rate law is;

-rA = ka[A][B]^2

2. α = 0, β = 1

-rA = ka[A]^0[B]

-rA = ka[B]

3. α=β=0

-rA = ka[A]^0[B]^0

-rA = ka

4. α=1, n = 0

The reaction is zero order as it is independent of the reactants.

-rA = ka

B. Rate law for the following;

a. H2+ Br2 >2HBr

-rA1 = ka[H2] . [Br2]

-rA2 = kb[HBr]^2

Comparing both rate, rA1= rA2

ka.[H2][Br2] = kb[HBr]^2

ka/kb = K = [HBr]^2 / [H2] [Br2]

b. H2 + I2 >2HI

K = [HI]^2 / [H2] [I2]

answer; talc and gypsum;

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.