Question 1 to 5 carries 1 mark each, from 6 to 9 carries 2 marks each, from 10 to 13 carries 3 marks each and question 14 carries 5 marks
Total Marks: 30 Duration: 1 hour
- The reaction $\ce{A + B -> C}$ has zero order. What is the rate of equation?
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Solution
\( Rate \propto [A][B] \)
- For a reaction $\ce{ A + H2O -> B }$ , \( Rate \propto [A] \). What is its (i) molecularity (ii) order of reaction?
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Solution
Order= 1, Molecularity= 2
- When is the rate of reaction equal to specific reaction rate?
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Solution
When concentration of each of the reactant is unity, the rate of reaction becomes specific rate.
- Why instantaneous rate is preferred over average rate?
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Solution
The rate of reaction at any time depends upon one of the reactants at that time which is not constant but goes on decreasing with time continuously, so that instantaneously rate gives more correct information than average rate.
- For the reaction $\ce{ 2X -> X2}$, the rate of reaction becomes three times when the concentration of X is increased 27 times. What is the order of reaction?
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Solution
3
- (a) The rate of reaction in the remaining mixture is not affected when a portion is removed for analysis of products, why? (b) Even an exothermic reaction has an activation energy, why?
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Solution
(a) Because the rate of reaction depends upon concentration which remains constant even if a portion is removed for analysis.
(b) Every reaction needs activation energy and it is required so that reactant can move together, can overcome forces of repultion and start breaking bond.
- (a) Give one example of a reaction having identical values of molecularity and order of reaction. (b) What factors determine whether a collision between two molecules will lead to a chemical reaction?
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Solution
(a) Decomposition of hydrogen peroxide
$\ce{ H2O2 -> H2O + {1 \over 2} O2 }$.
for this reaction order and molecularity is identical which is one.
(b) Temperature and shape of molecule is responsible for effective collision..
- Two reactions have identical values of \( E_{a}\). Dos this ensure that they will have same rate constant if run at the same temperature? Explain.
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Solution
No,according to Arhenius equation, \(k = Ae^{E_a \over RT }\) rate constant will depend on \( A, \, E_a \, and \, T \) as well.
- "The increase in reaction rate with rise in temperature is due to increase in the number of collisions." Is this statement correct? If not, correct then comment.
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Solution
Yes, this is correct according to collision theory, because with increase in temperature number of collision of molecules also get increase. Further, increase in concentration of reactant increase frequency of collision and so that increase reaction rate.
- For a chemical reaction $\ce{ A -> B }$, the change in concentration of A in 40s is -0.004 mol/litre. What is the rate of chemical reaction?
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Solution
Change in conc= \(- 0.004 \,mol\,L^{-1} \)
Change in time = 40 sec
\( Rate \, of \, chemical \, reaction = \frac{Change \, in \, conc}{Change \, in \, time} \)
\( = {-0.004 \, mol \, L^{-1} \over 40 sec} = 1 \times 10^{-4} \, mol\, L^{-1} \, s^{-1} \)
- For a reaction, the energy of activation is zero. What is the value of rate constant at 370 K, if \( k= 1.5 \times 10^4 \, s^{-1} \; at \, 370 \, K? \; R= 8.31 \, J\, K^{-1} \, mol^{-1} \)
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Solution
According to Arhenius equation
\( log \frac{k_2}{k_1} = {E_a \over 2.303 R}{ T_2 - T_1 \over T_1 T_2} \)
As \( E_a = 0 \,
and \, \frac{k_2}{k_1} = 1 \)
\(=> k_1= k_2 = 1.5 \times 10^4 s^{-1}\)
- Rate constant for first order reaction is \( 5.70 \times 10^{-5}\). What percentage of initial reaction will react in 5 hours?
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Solution
\( k= 5.70 \times 10^{-5}, \, t=5 \, h \)
Let initial conc = 100 , Change in conc = x
\( k={2.303 \over t} log {a \over a-x} \)
\( 5.70 \times 10^{-5} = \frac{2.303}{5 \times 3600} \times log \frac{100}{100-x} \)
on solving , we get \(x = 64.66 % \)
so percentage of initial concentration of amount reacted is 64.66%
- The pressure of a gas decomposing at the surface of a solid catalyst has been measured at different times and results are given below:
- (a) What is the difference between a unimolecular and a bimolecular elementry step? (b) If 1 % of the reactant decomposes in first minute in a first order reaction, how much of it will remain undecomposed after one hour?
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Solution
(a) Unimolecular elementary steps involves only one reactant and bimolecular elementary step involves two reactants.
(b) for the first order reaction
\( t = {2.303 \over k} log { a \over a-x } \)
initially, \( t = 1 min \, , \, k = ?, \, a = 100, \, a-x = 99 \)
then value of k is
\( k = {2.303 \over t} log { a \over a-x } \)
\( k = {2.303 \over 1} log { 100 \over 99 } \)
\( k = 0.010 min^{-1} \)
now to calculate the amount of reactant that would remain undecomposed
\( 60 min = {2.303 \over 0.010} log { 100 \over a-x } \)
\(log { 100 \over a-x } = 0.26 \)
\( a-x = 55.5 \)
So, 55.5% of reactant is remains undecomposed after 1 hour.
| Per second | 0 | 100 | 200 | 300 |
|---|---|---|---|---|
| Pressure (Pascals) | \( 4.0 \times 10^3 \) | \( 3.5 \times 10^3 \) | \( 1.0 \times 10^3 \) | \( 2.5 \times 10^3 \) |
Determine the order of reaction.
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Solution
This reaction is zero order
As it is zero order, then \( k = {1 \over t} \times ([A]_0-[A]) \)
in this case \( [A]_0 = 4.0 \times 10^3 \)
As the value of k is constant, therefore the reaction is of zero order and its rate constant is \( k= 5 \, Pa \, Sec^{-1} \)
Solution
This reaction is zero order
As it is zero order, then \( k = {1 \over t} \times ([A]_0-[A]) \)
in this case \( [A]_0 = 4.0 \times 10^3 \)
| Per second | Pressure (Pascals) | \( k = {1 \over t} \times ([A]_0-[A]) \) |
|---|---|---|
| 100 | \( 3.5 \times 10^3 \) | \( 5 \, Pa \, Sec^{-1} \) |
| 200 | \( 1.0 \times 10^3 \) | \( 5 \, Pa \, Sec^{-1} \) |
| 300 | \( 2.5 \times 10^3 \) | \( 5 \, Pa \, Sec^{-1} \) |
As the value of k is constant, therefore the reaction is of zero order and its rate constant is \( k= 5 \, Pa \, Sec^{-1} \)