Match the graph given in column I with the order of reaction given in column II. More than one item in Column I may link to the same item of Column II.
Column I
Column II
A
B
(I)
1st Order
C
(II)
Zero Order
D
Match the graph given in column I with the order of reaction given in column II. More than one item in Column I may link to the same item of Column II.
| Column I | Column II | ||
| A | |
||
| B | |
(I) | 1st Order |
| C | |
(II) | Zero Order |
| D | |
For first order reactions
A→ product
Rate law: Rate =k[A]1
Where, [A]= reactant concentration
On comparing this equation with
y=mx+c we get,
m(slope)=k
c(intercept)=0
Thus, the graph between rate and concentration will be linearly increasing.
(A)→(I)
For zero order reaction
Rate law expression:
Rate =k[A]∘
Rate =k
Rate = Constant
Thus, the graph between rate and concentration will be:
(B)→(II)
Integrated rate law for zero order
[A]t=[A]0−kt−−−−−−(a)
Where, [A]0= initial concentration of reactant
[A]t= reactant concentration at time ‘t′
k= rate constant
t= time
On comparing equation (a) with y=mx+c we get,
m(slope)=−k
c(intercept)=[A]0
Thus, the graph between concentration and time will be linearly decreasing with negative slope.
(C)→(II)
Integrated rate law for 1st order
k=1tln[A]0[A]t
ln[A]t[A]0=−kt
ln[A]t−ln[A]0=−kt
ln[A]t=−kt+ln[A]0
log[A]t=−kt2.303+log[A]0
[lnx=2.303logx]
On comparing this equation with
y=mx+c we get,
m(slope)=−k
c(intercept)=log[A]0
Thus, the graph between rate and concentration will be linearly decreasing with negative slope.
(D)→(I)
Hence, correct match is:
(A)→(I),(B)→(II),(C)→(II),(D)→(I)
For first order reactions
A→ product
Rate law: Rate =k[A]1
Where, [A]= reactant concentration
On comparing this equation with
y=mx+c we get,
m(slope)=k
c(intercept)=0
Thus, the graph between rate and concentration will be linearly increasing.
(A)→(I)
For zero order reaction
Rate law expression:
Rate =k[A]∘
Rate =k
Rate = Constant
Thus, the graph between rate and concentration will be:
(B)→(II)
Integrated rate law for zero order
[A]t=[A]0−kt−−−−−−(a)
Where, [A]0= initial concentration of reactant
[A]t= reactant concentration at time ‘t′
k= rate constant
t= time
On comparing equation (a) with y=mx+c we get,
m(slope)=−k
c(intercept)=[A]0
Thus, the graph between concentration and time will be linearly decreasing with negative slope.
(C)→(II)
Integrated rate law for 1st order
k=1tln[A]0[A]t
ln[A]t[A]0=−kt
ln[A]t−ln[A]0=−kt
ln[A]t=−kt+ln[A]0
log[A]t=−kt2.303+log[A]0
[lnx=2.303logx]
On comparing this equation with
y=mx+c we get,
m(slope)=−k
c(intercept)=log[A]0
Thus, the graph between rate and concentration will be linearly decreasing with negative slope.
(D)→(I)
Hence, correct match is:
(A)→(I),(B)→(II),(C)→(II),(D)→(I)
