Chemistry - Rates of Chemical Reactions and Chemical Kinetics
Exam Duration: 45 Mins Total Questions : 30
When spherical particles of 0.1 mm diameter are ground to spherical particles of 0.025 mm diameter (for the same volume), the surface area increase to
- (a)
4 times
- (b)
8 times
- (c)
16 times
- (d)
32 times
If the rate constant of a reaction is 2.0 mol-1 L-1 at 700 K and 32 mol-1 L s-1 at 800 K the activation energy for the reaction is
- (a)
120.11 kJ mol-1
- (b)
127.11 kJ mol-1
- (c)
129.11 kJ mol-1
- (d)
130.11 kJ mol-1
Which of the following best explains how catalysts increase the rate of a chemical reaction?
- (a)
They reduce the amount of product
- (b)
They increase the amount of the product
- (c)
They reduce the amount of reactants
- (d)
They reduce the activation energy for the recation
For the following reaction, which one you think will go with moderate rate
- (a)
\({ Cr }^{ 2+ }{ (aq) }{ + }{ Fe }^{ 3+ }(aq){ \longrightarrow }{ Cr }^{ 3+ }(aq)+{ Fe }^{ 2+ }(aq)\)
- (b)
\({ 2NO }_{ 2 }{ \longrightarrow }{ N }_{ 2 }{ O }_{ 4 }\)
- (c)
\({ 3Fe }^{ 2+ }(aq)+{ NO }_{ 3 }(aq)+{ 4H }^{ + }(aq)\longrightarrow 3{ Fe }^{ 3+ }(aq)+NO(g)+{ 2H }_{ 2 }O\)
- (d)
\({ C }_{ 8 }{ H }_{ 18 }+{ 12 }_{ 1/2 }{ O }_{ 2 }\longrightarrow { 8CO }_{ 2 }(g)+{ 9H }_{ 2 }O(g)\)
The activation energy for the reaction \(2HI(g)\quad \rightleftharpoons \quad { H }_{ 2 }(g)+{ I }_{ 2 }(g)\) is 209.5 kJ mol-1 at 581 K. The fraction of molecules of reactants having energy equal to or greater than activation energy would be
- (a)
1.462 x 1024
- (b)
1.462 X 10-19
- (c)
1.462 X 1028
- (d)
1.462 X 1034
Consider the chemical reaction, \({ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\longrightarrow { 2NH }_{ 3 }\) The rate of this reaction can be expresed in terms of time derivative of concentraction of N2(g), H(g) or NH3(g). Identify the correct relationship amongst the rate expressions.
- (a)
rate=\(-d\frac { \left[ { N }_{ 2 } \right] }{ dt } =-\frac { 1 }{ 3 } \frac { d\left[ { H }_{ 2 } \right] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } \)
- (b)
rate=\(-d\frac { \left[ { N }_{ 2 } \right] }{ dt } =-\frac { 3d\left[ { H }_{ 2 } \right] }{ dt } =\frac { 2d\left[ { NH }_{ 3 } \right] }{ dt } \)
- (c)
rate=\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } =-\frac { 1 }{ 3 } \frac { d\left[ { H }_{ 2 } \right] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } \)
- (d)
rate=\(-\frac { d\left[ { N }_{ 2 } \right] }{ dt } =-\frac { d\left[ { H }_{ 2 } \right] }{ dt } =\frac { d\left[ { NH }_{ 3 } \right] }{ dt } \)
99% of a first order reaction was completed in 32 min. When will 99.9% of the reaction complete?
- (a)
40 min
- (b)
46 min
- (c)
48 min
- (d)
52 min
The rate of reaction can be measured by nothing the change in some property of the reaction. The rate of reaction of which one of the following reactions can be studied by nothing down the pH change?
- (a)
\({ N }_{ 2 }{ O }_{ 4 }(g)\rightarrow 2{ NO }_{ 2 }(g)\)
- (b)
\({ N }_{ 2 }{ O }_{ 4 }(g)\rightarrow 2{ NO }_{ 2 }(g)+1/2{ O }_{ 2 }(g)\)
- (c)
\({ C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }\rightarrow { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }+{ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }\\ sucrose\quad \quad \quad \quad glucose\quad \quad fructose\)
- (d)
\({ CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }(aq)+{ H }_{ 2 }O(l)\rightarrow { CH }_{ 3 }COOH(aq)+{ C }_{ 2 }{ H }_{ 5 }OH(aq)\)
The reaction : A + 2B \(\rightarrow\) C+D obeys the rate equation. Rate = K [A]x [B]y
The overall order of this reaction is
- (a)
x
- (b)
y
- (c)
(x + y)
- (d)
zero
For the reaction A \(\rightarrow \) B, the rate law expression is rate = k[A]. Which of the following statements is incorrect?
- (a)
The reaction follows first order kinetics
- (b)
The t1/2 of reaction depends on initial concentration of reactants
- (c)
K is constant for the reaction at a constant temperature
- (d)
The rate law provides a simply way of predicting the concentration of the reactants and products at any time after the start of the reaction
The rate of a gaseous reaction is given by the expression k [A] [B]. If the volume of the reaction vessel is suddenly reduced to \(\frac { 1 }{ 4 } \)th of the original volume, the reaction rate relating to the original rate would be
- (a)
\(\frac { 1 }{ 10 } \)
- (b)
\(\frac { 1 }{ 8 } \)
- (c)
8
- (d)
16
A reaction \(A\rightarrow B\) follows a second order kinetics. Doubling the concentration of A will increase the rate of formation of B by a factor of
- (a)
2
- (b)
1/2
- (c)
4
- (d)
1/4
The chemical reaction \(2{ O }_{ 3 }\rightarrow 3{ O }_{ 2 }\) proceeds as follows:
\({ O }_{ 3 }\rightleftharpoons { O }_{ 2 }+O\) (fast)
\(O+{ O }_{ 3 }\rightarrow 2{ O }_{ 2 }\) (slow)
The rate law expression should be
- (a)
r = k [O2] [O3]
- (b)
r = k [O3]2
- (c)
r = k [O3]2[O2]-1
- (d)
NONE OF THE ABOVE
A plot of the concentration of A versus time found for the reaction A \(\rightarrow \) products, is a astraight line. The order of the reaction is
- (a)
zero
- (b)
one
- (c)
two
- (d)
0.5
Which of the following statements is correct?
- (a)
For reaction xX \(\rightarrow \) yY
Rate = \(\frac { dX }{ xdt } \)=\(\frac { dY }{ dt } \)
- (b)
Ea = ER + Ethreshold
- (c)
The parameter, rate constant and specific reaction rate have different meanings.
- (d)
For any reaction the value of specific rate is independent of the initial concentration of reactants
The reaction:
\({ N }_{ 2 }{ O }_{ 5 }(in\quad CC{ l }_{ 4 })\rightarrow 2N{ O }_{ 2 }(g)+\frac { 1 }{ 2 } { O }_{ 2 }(g)\)
is first order in N2O5 with rate constant 6.2X10-4s-1. What is the value of rate of reaction when [ N2O5] = 1.25 mol L-1?
- (a)
7.75 X 10-4 mol L-1 s-1
- (b)
6.65 X 10-3 mol L-1 s-1
- (c)
5.15 X 10-5 mol L-1 s-1
- (d)
3.85 X 10-4 mol L-1 s-1
For the reaction: \({ A\rightarrow B }\), the rate law expression is: rate = k[A]. Which of the following statements is incorrect?
- (a)
The reaction follows first order kinetics.
- (b)
The t1/2 of teh reaction depends on initial concentration of reactants
- (c)
k is constant for the reaction at constant temperature
- (d)
The rate law provides a simple way of predicting the concentration of reactants and products at any time after the start of the reaction.
How will be the rate of reaction
\(2S{ O }_{ 2 }(g)+{ O }_{ 2 }(g)\rightarrow 2S{ O }_{ 3 }(g)\)
change if the volume at the reaction vessel is halved
- (a)
It will be 4 times of its initial value
- (b)
It will be 8 times of its initial value
- (c)
It will be 1/4 of its initial value
- (d)
It will be 1/6th of its initial value
Consider the following reaction, \(2H_2O+O_2\longrightarrow2H_2O\)The rate law expression, r=k[H2]n for the above reaction.When the concentration of H2 is doubled, the rate of reaction found to be quadrupled.The value of n is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
For a unimolecular reaction
- (a)
the molecularity and order of a reaction is one
- (b)
the molecularity of the reaction is one while their order is zero
- (c)
two reacting species are involved in the rate determining
- (d)
the molecularity and order of the slowest step of the reaction is equal to one
A first order is 10% complete in 20 min.The time taken for 19% completion is
- (a)
30min
- (b)
38min
- (c)
40min
- (d)
45min
Consider the following reaction, \(5Br^-(aq)+BrO^{-}_{3}(aq)+6H^+(aq)\longrightarrow3Br_2(aq)+3H_2O(l)\) Which one of the following expression is correct for the rate of reaction for the above chemical equation?
- (a)
\({\Delta[Br^-]\over\Delta t}=6{\Delta[H^+]\over\Delta t}\)
- (b)
\({\Delta[Br^-]\over\Delta t}={6\over5}{\Delta[H^+]\over\Delta t}\)
- (c)
\({\Delta[Br^-]\over\Delta t}=5{\Delta[H^+]\over\Delta t}\)
- (d)
\({\Delta[Br^-]\over\Delta t}={5\over6}{\Delta[H^+]\over\Delta t}\)
The rate of formation of SO3 in the following reaction is 100g min-1 \(2SO_2+O_2\longrightarrow2SO_3\)The rate of disappearance of O2 is
- (a)
29 min-1
- (b)
20g min-1
- (c)
50g min-1
- (d)
200g min-1
Consider the following reaction, \(2A+B+C\longrightarrow\) products How will the rate of reaction changes when the concentration of A is doubled and that B is tripled while C is taken in excess?
- (a)
The rate reduces 8 times of its original value
- (b)
The rate reduces 12 times of its original value
- (c)
The rate of increase 8 times of its original value
- (d)
The rate of increases 12 times of its original value
The rate of a reaction double when its temperature changes from 300k to reaction will be9R=3.314JK-1 mol-1 and log2=0.301)
- (a)
53.6KJ mol-1
- (b)
48.6KJ mol-1
- (c)
58.5KJ mol-1
- (d)
60.5KJ mol-1
the time for half-life period of certain reaction, A \(\rightarrow\) products is 1 h.When the initial concentration of the reactant 'A', is 2.0 mol L-1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L-1, if it is a zero order reaction?
- (a)
4h
- (b)
0.5h
- (c)
0.25h
- (d)
1h
For a reaction, \({1\over2}A\longrightarrow2B,\) rate of disappearance of 'A' is related of the rate of appearance of 'B' by the expression
- (a)
\(-{d[A]\over dt}={1\over2}{d[B]\over dt}\)
- (b)
\(-{d[A]\over dt}={1\over4}{d[B]\over dt}\)
- (c)
\(-{d[A]\over dt}={d[B]\over dt}\)
- (d)
\(-{d[A]\over dt}=4{d[B]\over dt}\)
The energies of activation for forward and reverse reaction for \(A_2+B_2\rightleftharpoons2AB\) are 180kJ mol-1 and 200kJ mol-1 respectively.The presence of a catalyst lowers the activation energy of both reactions by 100 kJ mol-1.The enthalpy change of the reaction (A2+B2\(\rightarrow\)2AB) in the presence of catalyst will be (in kJ mol-1)
- (a)
300
- (b)
120
- (c)
280
- (d)
20
The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr
\(NO(g)+Br_2(g)\rightleftharpoons NOBr_2(g)\)
\(NOBr_2(g)+NO(g)\longrightarrow 2NOBr(g)\)
If the second step is the rate determining step, the order of the reaction with respect to NO(g) is
- (a)
1
- (b)
0
- (c)
3
- (d)
2
A reaction involving two different reactants can never be
- (a)
bimolecular reaction
- (b)
second order reaction
- (c)
first order reaction
- (d)
unimolecular reaction
1654060945.png)
1654060942.png)
1653987551.png)
1653987086.png)
1582095229.jpg)
