1) In the reaction A-->2B, the rate law was determined to be rate=k[A] 2 (k=0.1
ID: 871961 • Letter: 1
Question
1) In the reaction A-->2B, the rate law was determined to be rate=k[A]2 (k=0.1 M-1min-1). When the initial [A]=0.5M, what is the concentration of A after one hour?
2) The half life of 131I is 8 days. How long will it take for 99.9% of a sample containing 131I to decay?
3) reaction A-->2B is first order in A. At time=0s, [A] was 1.0M and [B] was 0M. After 100s, [B] increased to 0.5M. What is the half life of A in this reaction?
4) The first order reaction A-->B+C has a half life (t1/2) of 30 min at 300K and 3min at 400K. What is the activation energy of the reaction? Can somebody show me how to do it step by steps!
Explanation / Answer
1)The rate constant units suggest the reaction to be second order
The rate of reaction -dCa/dt= 2KCa2 ( assuming constant volume and liquid phase reaction)
UPon integration
1/Ca-1/Cao= 2kt
K=0.1 , t= 60minutes= 3600sec
1/Ca-1/0.5= 2*0.1*60*60
1/ca =1/0.5+3600
Ca=0.000278M
2) half life is given by =0.693/ K =where K is rate constant
K =0.693/ half life =0.693/8=0.086625/days
for first order reaction
ln(Cao/Ca)= Kt (1)
Ca=concentration at time t
Percent decayed X=0.999
X=1-Ca/Cao= Ca/Cao=1-0.999=0.001
Cao/Ca=1000
Substituting in (1)
ln1000= 0.08625t
t=ln1000/0.08625=79.74 days
3) As per stoichiometry, for every 1 moles of A reacted, 2 moles of B is formed
Since concentraton ogf B is 0.5M, 0.25M of A must have reacted in 100 seconds
Cao=1M
Ca=0.75M
t=100second
For first order reaction
ln(Cao/Ca)=kt
ln(1/0.75)=1000K
K=0.00287/sec
Half life= 0.693/K= 0.693/0.00287=240 sec
4) For first order reaction
K=0.693/half life
at 300K
K300=0.693/30=0.0231/min
at 400K
K400 =0.693/3 =0.231/min
From Arhenius equation
ln(K400/K300)= (Ea/R)*(1/300-1/400)
ln(0.231/0.0231)= (Ea/R) *0.000833
2.3= (Ea/R)*0.000833
Ea/R= 2/0.000833=2400
Ea=2400*8.314 J/molK=19953.6 J/mol
ln(1000)
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