1821 fo 178 The decomposition of HI to H2 12 at 508 has a h

1821 fo [178] The decomposition of HI to H2 + 12 at 508 has a half-life of 135 min when the initial, pressure of HI is 0.1 atm and 13.5 min when the pressure is 1 atm. ) Show that this proves that the reaction is second order. What is the value of the rate constant in Lmol\'s and bar\" s (I)H1 2nd-reaction,4,2 -kpP40 , PA,/2-h (135)(0.1)-(13.5)(1.0)-constant 1--= 04/2-(0. 1x 1.01 325)(135x60) (2) (A)k, = x1.01325)135x60)121810\'bar\'s kC = [(1.21 8x1 0-3)(1.01 325)(0.08205 x 781)-(1-2)0.079 Lrnole is |[1791C f i amnle of a gaseous compound initially at 3

Solution

a) The problem starts with the assumption that the reaction is indeed 2^nd order and then sets out to prove that it really is 2^nd. order.

The integrated rate law for a 2^nd order reaction is given as

1/[A]_t = 1/[A]₀ + k*t

where k is the 2^nd order rate constant; t is the time taken; [A]_t is the concentration of the reactants at time t and [A]₀ is the initial concentration.

Since, we are dealing with a gas phase reaction, we can replace [A]_t by P_t and [A]₀ by P₀ where P₀ and P_t are the partial pressures of the gas. The half life of the reaction, t_1/2 is defined as the time taken for the reactant to reduce to one half of its initial concentration (or initial pressure). Therefore,

P_t = ½*P₀; therefore,

1/(P₀/2) = 1/P₀ + k*t_1/2

=====> 2/P₀ = 1/P₀ + k*t_1/2

=====> 1/P₀ = k*t_1/2

=====> t_1/2 = 1/k.P₀ ………(1)

Given P₀ = 0.1 atm; t_1/2 = 135 min, we have, k = 1/(0.1 atm).(135 min) = 1/(13.5 atm-min).

Again, when P₀ = 1 atm; t_1/2 = 13.5 min, we will have k = 1/(1.0 atm).(13.5 min) = 1/(13.5 atm-min).

The right hand side of both are equal and is equal to the 2^nd order rate constant; therefore, the reaction is indeed 2^nd order (ans).