Question #20230

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

Expert's answer

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 %)

t(1/2) = ln2/K

K = ln2/t(1/2)

K = ln2/ 5 x 10^4 = 1.38 x 10^-5

ln(C

t = 7200 s

ln(C

C

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