Given that the equation for a first-order rate law is in((A)_t/(A)_o)=-kt, pleas
ID: 987985 • Letter: G
Question
Given that the equation for a first-order rate law is in((A)_t/(A)_o)=-kt, please answer the following questions: 2H_2O_2(aq) right arrow 2H_2O_omega + O_2_(g) The reaction is first-order with respect to H_20_2, the rate constant for the consumption of H_20_2 at 20 degree C is 1.8 times 10^5S^-1, and the initial concentration of H_2O_2 is 0.30 M. What is the concentration of H_2O_2 after 4h? How long will it take the H_2O_2 concentration to drop to 0.12 M? How long will it take for 90% of the H_20_2 to decompose?Explanation / Answer
If first order then
lnA = lnA0 - kt
k = 1.8*10^-5
A0 = 0.3
a)
aftetr t = 4 hours
note that the k is givne in seconds so change either t or k to the same units
4 hours = 4*3600 = 14400 seconds
then
lnA = ln(0.3) - (1.8*10^-5)(14400)
lnA = -1.463172
A = exp(-1.463172) = 0.23150 M
b)
t for A = 0.12
substitute in
lnA = lnA0 - kt
ln(0.12) = ln(0.3) - (1.8*10^-5) * t
t = (ln(0.12/0.3))/(-1.8*10^-5) = 50905.04 seconds
t = 14.14 hours
c)
time required fo 90% of H2O2 to ddecompose
if 90% decomposes, then there is only 100-90 = 10% left so A = 0.1A0
ln(0.1*A0) = ln(A0) - (1.8*10^-5) * t
ln(0.1*A0/A0) = - (1.8*10^-5) * t
t = ln(0.1)/(- (1.8*10^-5) ) = 127921.394055 seconds or
t = 35.533 hours
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