Answer to Question #20230 in Physical Chemistry for Abhishek

Question

27. The half life for a first order reaction is 5 x 104 s. What percentage of the initial reactant
will react in 2 hours? Calculate

Answer 1

t(1/2) = 5 x 10^4 - half life for a first orderreaction

t(1/2) = ln2/K
K = ln2/t(1/2)
K = ln2/ 5 x 10^4 = 1.38 x 10^-5

ln(C_x/C_o) = -Kt

t = 7200 s

ln(C_x/C_o) = -(1.38 x 10^-5 * 7200)

C_x/C_o = 0.905 if C_o = 1, x = 0.905 (9.05 %)