(a)
Interpretation:
An nth order in a reaction A is given by an equation t1/2 has to be shown
Concept introduction:
Rate law for the nth order is given below,
d[A]dt = −kr[A]n
(a)
Explanation of Solution
Rate law for the nth order is given below,
d[A]dt = −kr[A]n
Here,
Initial concentration of reactant is [A].
Rate constant of the reaction is kr.
The reaction is rearranged,
d[A]dt = −kr[A]n
The reaction is integrated,
-[A]∫[A0]d[A][A]n = kr t∫0dt-11-n(1[A]n-1)[A][A0] = krt-11-n(1[A]n-1- 1[A]n-10)[A][A0] = krt [A]1-n- [A]1-n0= krt(n-1)
An nth order in a reaction A is given by an equation t1/2 is given.
(b)
Interpretation:
Time for the concentration of a substance to fall to one·third the initial value in an nth·order reaction has to be derived.
Concept introduction:
Balancing reaction:
Refer to part (a)
(b)
Explanation of Solution
Half-life of the reaction is t1/2.
Half-life of the nth order reaction is given below,
Lets,
[A] =0.5 [A]0
Therefore,
[A]1-n- [A]1-n0= krt(n-1) [0.5A]1-n0- [A]1-n0= krt1/2(n-1) [A]1-n0(0.51-n -1)= krt1/2(n-1) [A]1-n0((2-1)1-n -1)= krt1/2(n-1)
t1/2 = [A]1-n0((2-1)1-n -1)kr(n-1)t1/2 = 2n-1-1[A]1-n0kr(n-1)t1/2 = 2n-1-1(n-1)kr[A]n-10
Time is calculated when the concentration of a substance to fall to one third the initial value in an nth·order reaction.
[A]1-n- [A]1-n0= krt(n-1) [(13)A]1-n0- [A]1-n0= krt(n-1) 3n−1[A]1-n0- [A]1-n0= krt(n-1) [A]1-n0(3n−1 -1)= krt1/2(n-1)t = [A]1-n0(3n−1 -1)kr(n-1)t1/2 = 3n-1-1(n-1)kr[A]n-10
Time for the concentration of a substance to fall to one·third the initial value in an nth·order reaction is shown.
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Chapter 17 Solutions
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