The rate of a first order reaction is 0.04mol per litre per sec at 10 minutes and 0.03 mol per litre per sec at 20 minutes after initiation. Find the half life of the reaction?

for the 1st order reaction rate constant k is given by

k =( 2.303/t₂-t₁) log([R₁]/[R₂])

according to the above problem

t1 = 10 min

t2 = 20 min

therefore t2-t1 = 10 min

also,

[R1] = 0.04 mol/L

[R2] = 0.03 mol/L

substituting these values in the above equation

k =( 2.303/10) log( 0.04/0.03)

k = 0.2303 [log(0.04) - log (0.03)]

k = 0.2303 [-1.3979 - (-1.5228)]

k = 0.02876 per min

for the first order reaction half life t_1/2 is given by

t_1/2 = 0.693/k

putting k = 0.02876 we get

t_1/2 = 0.693/0.02876

t_1/2 = 24.096 min