The reactant concentration in a zero-order reaction was 0.100 M after 160 s and
ID: 887972 • Letter: T
Question
The reactant concentration in a zero-order reaction was 0.100 M after 160 s and 2.50×102M after 365 s . What is the rate constant for this reaction?
Correct
Part B
What was the initial reactant concentration for the reaction described in Part A?
.0586
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Part C
The reactant concentration in a first-order reaction was 7.20×102M after 30.0 s and 4.20×103Mafter 100 s . What is the rate constant for this reaction?
Part D
The reactant concentration in a second-order reaction was 0.750 M after 295 s and 6.20×102M after 815 s . What is the rate constant for this reaction?
k0th = 3.66×104 M/sExplanation / Answer
PART A:
reactant concentration in a zero-order reaction was(a-x)1= 0.100 M (a-x)2= 2.50x10-2M
after time t1=160 s t2= 365 s
k0=1/(t2-t1)[(a-x)2-(a-x)1]
= (1/205)[2.5x10-2 -0.100]
= 0.075/205
K0 = 3.65x10-4
PART B
was the initial reactant concentration for the reaction described in Part A?
K0 = 3.65x10-4 (a-x)= 0.100 M a=?
k0=1/(t0)[(a--(a-x)]
3.65x10-4 =[1/160]x[a-(0.100)]
3.65x10-4x160 =[a-(0.100)]
584x10-4=(a-0.100)
584x10-4 =[a-(0.100)]
a= 0.0584+0.100
a=0.1584
PART C
(a-x)1= 7.20x10-2 (a-x)2= 4.20x10-3 t1=30 t2 =100s
K1=2.303/(t2-t1)xlog[(a-x)1/(a-x)2]
k1= 2.303/(100-30)xlog[(7.20x10-2)1/(4.20x10-3)]
k1= [2.303/(70)]xlog[(7.20x10-2)/(4.20x10-3)]
k1=0.0329xlog[17.14]
k1=0.0329x1.2340
k1=0.04059
k1=4.05x10-2 S-1
PART D:
(a-x)1= 0.750 M t1= 295 s(a-x)2= 6.20×102M t2=815 s . What is the rate constant for this reaction?
K2=1/(t2-t1) [1/(a-x)2-(1/a-x)1]
k2= [1/(815-295)]x[(1/ 6.20x10-2)-(7.50x10-2)]
k2=[1/(520)]x[(7.50x10-2)/ 6.20x10-2)-(6.20x10-2/7.50x10-2)]
k2=[1.92x10-3]x[(1.3x10-2)/46.5x10-4]
k2=[1.92x10-3)]x[0.0279x102]
k2=[1.92x10-3)]x[0.0279x102]
k2=5.367x10-3M1s1
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