IISER Chemistry - Rates of Chemical Reactions and Chemical Kinetics
Exam Duration: 45 Mins Total Questions : 30
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
When concentractions are expressed in mole units and time in seconds, the units of the zero order rate constant (k) are
- (a)
mol L-1 s-1
- (b)
L mol-1 s-1
- (c)
s-1
- (d)
L2 mol-2 s-1
A reaction that is of first order with respect to reactant A has a rate constant 6 mim-1. If we start with [A] = 0.5 mol L-1, when would [A] reach the value 0.05 mol L-1?
- (a)
3.84 min
- (b)
3 min
- (c)
0.15 min
- (d)
0.384 min
For the reaction: Cl2+2I- \(\longrightarrow \)2Cl-+I2 the initial concentraction of iodide ions (I-) was 0.5 mol L-1 and that after 10 minutes was 0.46 mol L-1. The rate of disappearance of I-, and the rate of apperance of iodine(I2) will be respectively in mol L-1 min-1.
- (a)
0.004 and 0.002
- (b)
0.004 and 0.003
- (c)
0.004 and 0.001
- (d)
0.0002 and 0.0002
For the reaction \({ H }_{ 2 }{ O }_{ 2 }(aq)+{ 2H }^{ + }(aq)+{ 2I }^{ - }(aq)\longrightarrow { I }_{ 2 }+{ 2H }_{ 2 }O\) rate is given by the expression k [H2O2][I-(aq)]2 the overall order to the reaction is
- (a)
one
- (b)
two
- (c)
three
- (d)
four
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=\(-\frac { d\left[ { N }_{ 2 } \right] }{ dt } =-\frac { d\left[ { H }_{ 2 } \right] }{ dt } =\frac { d\left[ { NH }_{ 3 } \right] }{ dt } \)
The decomposition of A follows second order kinetics.The initial rate of decomposition when [A]0=0.50 M is
- (a)
the same as the initial rate for any value of (A)0
- (b)
half of that found when (A)0=1.00 M
- (c)
five times the rate found when (A)0=0.10 M
- (d)
four times the rate found when (A)0=0.25 M
Oxidation of oxalic acid by acidified \(KMnO_4\) is an example of auto catalysis; it is due to which of the following :
- (a)
SO42-
- (b)
MnO42-
- (c)
Mn2+
- (d)
K+
The half life period for catalytic decomposition of \(AB_3\) at 50 mm is found to be 4 hours and at 100 mm it is 2.0 hours. The order of reaction is
- (a)
1
- (b)
2
- (c)
3
- (d)
0
A chemical reaction involved two reacting species. The rate reaction is directly proportional to the concentraction of one of them and inversly proportional to the concentration of the other. The order of reaction is
- (a)
zero
- (b)
1
- (c)
2
- (d)
unpredictable
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
The rate expression for the reaction
aA \(\rightarrow\) Product, would be
- (a)
\(\frac { -d[A] }{ dt } =k{ [A] }^{ a }\)
- (b)
\(\frac { -d[A] }{ dt } =k{ [A] }^{ 0 }\)
- (c)
\(\frac { -d[A] }{ dt } =k{ [A] }^{ x }\)
- (d)
\(\frac { -d[A] }{ dt } =k{ [A] }^{ 1 }\)
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
In a catalytic conversion of N2 to NH3 by Haber's process, the rate of reaction expressed as change in the concentration of ammonia per times is 40 X 10-13 mol-1s-1. If there are no side reactions, the rate of the reaction as expressed in terms of hydrogen is
- (a)
60 X 10-3 mol-1 s-1
- (b)
20 X 10-3 mol-1 s-1
- (c)
30 X 10-3 mol-1 s-1
- (d)
10 X 10-3 mol-1 s-1
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
For the chemical reaction the following rate equation
\(\frac { d[A] }{ dt } =k[A]^{2}\)
The units for rate constant is
- (a)
time
- (b)
minute-1
- (c)
sec-1mol-1litre+1
- (d)
minute-1mol-1litre-1
The Arrhenius equation expressing the effect of temperature on the rate constant of the reaction is
- (a)
k = e-Ea/RT
- (b)
\(k=\frac { { E }{ a } }{ RT } \)
- (c)
\(k=lo{ g }_{ e }\frac { { E }{ a } }{ RT } \)
- (d)
k = Ae-Ea/RT
The reaction \(2{ N }_{ 2 }{ O }_{ 5 }(g)\rightarrow 4N{ O }_{ 2 }(g)+{ O }_{ 2 }(g)\) is first order w.r.t. N2O5. Which of the following graphs would yield a straight line?
- (a)
Log p N2O5 vs. time with a -ve slope
- (b)
(p N2O5)-1 vs. time
- (c)
p N2O5 vs. time
- (d)
log p N2O5 vs time with a +ve slope.
The differential rate law for reaction:
\({ H }_{ 2 }+{ I }_{ 2 }\rightarrow 2HI\) is
- (a)
\(-\frac { d[{ H }_{ 2 }] }{ dt } =-\frac { d[{ I }_{ 2 }] }{ dt } =-\frac { d[HI] }{ dt } \)
- (b)
\(\frac { d[{ H }_{ 2 }] }{ dt } =\frac { d[HI] }{ dt } =\frac { 1 }{ 2 } \frac { d[HI] }{ dt } \)
- (c)
\(\frac { 1 }{ 2 } \frac { d[{ H }_{ 2 }] }{ dt } =\frac { 1 }{ 2 } .\frac { d[{ I }_{ 2 }] }{ dt } =-\frac { d[HI] }{ dt } \)
- (d)
\(-\frac { 2d[HI] }{ dt } =\frac { -2d[{ I }_{ 2 }] }{ dt } =\frac { d[HI] }{ dt } \)
Consider the following reaction, \(N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)\)The rate of change of concentration of ammonia is
- (a)
0.2x10-4Ms-1
- (b)
0.4x10-4Ms-1
- (c)
0.6x10-4Ms-1
- (d)
-0.6x10-4Ms-1
For the reaction, R\(\longrightarrow\)P, the concentration of a reactant changes from 0.03M to 0.02M in 25 min.Calculate the average rate of reaction using units of time in seconds.
- (a)
6.66x10-5
- (b)
6.6x10-6
- (c)
5.67x10-5
- (d)
7.26x10-6
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
Compounds 'A' and 'B' react according to the following chemical equation, \(A(g)+2B(g)\longrightarrow2C(g)\)Concentration of either 'A' or 'B' were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration.Following results were obtained.Choose the correct option for this reaction.
| Experiment | Initial com. of [A]/mol L-1 | Initial com. of [A]/mol L-1 | Initial rate of formation of [CV]/mol L-1s-1 |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
- (a)
Rate=k[A]2[B]
- (b)
Rate=k[A][B]2
- (c)
Rate=k[A][B]
- (d)
Rate=k[A]2[B]0
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
Consider the following reaction, \(2N_2O_5\longrightarrow4NO_2+O_2+{d[NO_2]\over dt}=k_2[N_2O_5]\)\({d[O_2]\over dt}=k_3[N_2O_5],{d\over dE}[N_2O_5]=k_1\)The relation between k1,k2 and k3 is
- (a)
k1=k2=k3
- (b)
2k1=k2=4k3
- (c)
2k1=4k2=k3
- (d)
None of these
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
For a first order reaction, (A)\(\rightarrow\)products, the concentration of A change from 0.1M of reaction when concentration of A 0.01M will
- (a)
1.73x10-5M min-1
- (b)
2.47x10-4M min-1
- (c)
3.47x10-4M min-1
- (d)
1.75x10-4M min-1
The rate of a chemical reaction double for every 100C rise of temperature.If the temperature is raised by 50 0, the rate of reaction increase by about
- (a)
10times
- (b)
24 times
- (c)
32 times
- (d)
64 times
The half-life period of a first order chemical reaction is 6.93min.The time required for the completion of 99% of chemical reaction will be
- (a)
230.6 min
- (b)
23.03 min
- (c)
46.06 min
- (d)
460.6 min
For the non-stoichiometric reaction \(2A+B\rightarrow C+D\), the following kinetic data were obtained in three separate experiments, all at 298K.
| Initial concentration |
Initial Concentration |
Initial rate of formation of C (mol L-1S-1) |
| 0.1M | 0.1M | 1.2x10-3 |
| 0.1M | 0.2M | 1.2x10-3 |
| 0.2M | 0.1M | 1.4x10-3 |
The rate law for the formation of C is
- (a)
\({dC\over dt}=k[A][B]\)
- (b)
\({dC\over dt}=k[A]^2[B]\)
- (c)
\({dC\over dt}=k[A][B]^2\)
- (d)
\({dC\over dt}=k[A]\)
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