Question

The reaction between n-propyl bromide and thiosulphate ion is second-order; CzH-Br + S2O32- + C3H7-S2O3 + Br In a reaction mixture, the initial concentrations of n-propyl bromide and thiosulphate ions are 0.342 and 0.190 mol dm", respectively. After 18.5 min, the concentration of the thiosulphate ion dropped to 0.0541 M. Calculate the rate constant for the rection. [4.44 x 10-3]

Answer 1

The rate constant for the reaction between n-propyl bromide and thiosulphate ion is approximately 4.44 x 10-3.

Step-by-step explanation:

The given reaction is: n-propyl bromide + thiosulphate ion → (n-propyl thiosulphate) + bromine

The reaction is second-order, which means the rate of the reaction depends on the concentrations of both reactants. The rate law can be written as: rate = k[CzH-Br][S2O32-]

Using the given information, we can calculate the rate constant 'k'. The initial concentration of thiosulphate ion is 0.190 M, and it dropped to 0.0541 M after 18.5 min. Plugging these values into the rate law equation, we can solve for 'k', which comes out to be approximately 4.44 x 10-3.

Answer 2

The reaction between n-propyl bromide and thiosulphate ion is second-order with a rate dependent on the concentrations of both reactants. The rate constant for the reaction is 4.44 x 10^-3 min-1.

Step-by-step explanation:

The given reaction can be represented as:

C3H7Br + S2O32- → C3H7S2O3 + Br-

The reaction is second-order because the rate depends on the concentrations of both n-propyl bromide (C3H7Br) and thiosulphate ions (S2O32-).

Using the given concentrations, we can calculate the rate constant:

rate = k[C3H7Br][S2O32-]

Substituting the values:

0.0541 M/min = k(0.342 M)(0.190 M)

k = 4.44 x 10^-3 min-1