Home
MCQ
TNPSC
TNPSC Group 1
TNPSC Group 2 2A
TNPSC Group 4 VAO
UPSC
TNTET
TNTET Paper 1
TNTET Paper 2
TNUSRB
TNUSRB PC
TNUSRB SI
Defence (NDA,CDS,AFCAT)
NDA
CDS
AFCAT
NEET
SSC
SSC CGL
SSC CHSL
SSC MTS
SSC GD
RRB
RRB NTPC
RRB Group D
RRB ALP
RRB JE
Blog
Reach Us
Login
The Effect of Temperature on the Rate of Reactions NEET Questions
NEET SYLLABUS
Physical Chemistry - Chemical Kinetics
Rate of a Chemical Reaction and Factors Affecting the Rate of Reactions (Concentration, Temperature,... )
Order and Molecularity of Reactions
Rate Law
Rate Constant and its Units
Differential and Integral Forms of Zero and First-Order Reactions and their Characteristics and Half-Lives
The Effect of Temperature on the Rate of Reactions
Collision Theory of Bimolecular Gaseous Reactions
Arrhenius Equation
Activation Energy and its Calculation
The Effect of Temperature on the Rate of Reactions MCQ Questions
Prev
1
2
3
4
5
6
7
8
9
10
Next
1-10
7.
The plot of ln k versus 1/T according to the Arrhenius equation is:
A.
a parabola
B.
a straight line with positive slope
C.
a curve passing through origin
D.
a straight line with negative slope
😑
View Answer
Rough Work
Error
ANSWER
:
D. a straight line with negative slope
8.
The slope of the straight line obtained by plotting ln k against 1/T is equal to:
A.
–Ea/R
B.
–Ea/2.303 R
C.
+Ea/R
D.
–R/Ea
😑
View Answer
Rough Work
Error
ANSWER
:
A. –Ea/R
9.
The slope of the line obtained by plotting log k versus 1/T is:
A.
+Ea / (2.303 R)
B.
–2.303 R / Ea
C.
–Ea / R
D.
–Ea / (2.303 R)
😑
View Answer
Rough Work
Error
ANSWER
:
D. –Ea / (2.303 R)
10.
The intercept of a plot of ln k versus 1/T (Arrhenius plot) is:
A.
ln Ea
B.
–Ea/R
C.
log A
D.
ln A
😑
View Answer
Rough Work
Error
ANSWER
:
D. ln A
11.
The integrated form of the Arrhenius equation relating rate constants k1 and k2 at temperatures T1 and T2 is:
A.
log(k2/k1) = (Ea/2.303 R) · (T2 – T1)/(T1·T2)
B.
log(k2/k1) = (Ea/2.303 R) · (T1·T2)/(T2 – T1)
C.
log(k2/k1) = (R/2.303 Ea) · (T2 – T1)/(T1·T2)
D.
log(k2/k1) = (Ea/R) · (T2 – T1)
😑
View Answer
Rough Work
Error
ANSWER
:
A. log(k2/k1) = (Ea/2.303 R) · (T2 – T1)/(T1·T2)
12.
The rate of a reaction doubles when its temperature changes from 300 K to 310 K. The activation energy of the reaction is approximately (R = 8.314 J K⁻¹ mol⁻¹):
A.
26.8 kJ mol⁻¹
B.
53.6 kJ mol⁻¹
C.
107 kJ mol⁻¹
D.
13.4 kJ mol⁻¹
😑
View Answer
Rough Work
Error
ANSWER
:
B. 53.6 kJ mol⁻¹
Prev
1
2
3
4
5
6
7
8
9
10
Next
1-10
Your Name
*
Your Email
*
Justify your answer :
*
Send Message