Chemical Kinetics Test

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Chemical Kinetics Test - 79

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Question 1
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Directions For Questions

The following Rice-Herzfeld mechanism has been suggested for the formation of phosgene $$CO{Cl}_{2}$$:
(1) $${ Cl }_{ 2 }\longrightarrow 2Cl\quad \left[ { K }_{ 1 }={ A }_{ 1 }.{ e }^{ -{ E }_{ { a }_{ 1 } }/RT } \right] $$
2. $$2Cl\longrightarrow { Cl }_{ 2 }\quad \left[ { K }_{ 2 }={ A }_{ 2 }.{ e }^{ -{ E }_{ { a }_{ 2 } }/RT } \right] $$
3. $$Cl+CO\longrightarrow COCl\quad \left[ { K }_{ 3 }={ A }_{ 3 }.{ e }^{ -{ E }_{ { a }_{ 3 } }/RT } \right] $$
4. $$COCl\longrightarrow Cl+CO\quad \left[ { K }_{ 4 }={ A }_{ 4 }.{ e }^{ -{ E }_{ { a }_{ 4 } }/RT } \right] $$
5. $$COCl+{ Cl }_{ 2 }\longrightarrow COCl +Cl\quad \left[ { K }_{ 5 }={ A }_{ 5 }.{ e }^{ -{ E }_{ { a }_{ 5 } }/RT } \right] $$

...view full instructions

What is overall activation energy of reaction, if steps (1) to (4) are much faster than (5)?

A
$$\cfrac { { E }_{ { a }_{ 5 } }.{ E }_{ { a }_{ 3 } }.{ E }_{ { a }_{ 1 } }^{ 1/2 } }{ { E }_{ { a }_{ 4 } }.{ E }_{ { a }_{ 2 } }^{ 1/2 } } $$

B
$${ E }_{ { a }_{ 5 } }+{ E }_{ { a }_{ 3 } }+\cfrac { 1 }{ 2 } { E }_{ { a }_{ 1 } }-{ E }_{ { a }_{ 2 } }- { E }_{ { a }_{ 4 } }$$

C
$${ E }_{ { a }_{ 1 } }+{ E }_{ { a }_{ 3 } }+{ E }_{ { a }_{ 5 } }-{ E }_{ { a }_{ 2 } }-{ E }_{ { a }_{ 4 } }\quad $$

D
$$-{ E }_{ { a }_{ 5 } }-{ E }_{ { a }_{ 3 } }-\cfrac { 1 }{ 2 } { E }_{ { a }_{ 1 } }+{ E }_{ { a }_{ 2 } }+\cfrac { 1 }{ 2 } { E }_{ { a }_{ 4 } }$$
Question 2
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Arrhenius equation gives the change in rate constant (and hence rate of reaction) with temperature. If the activation energy of the reaction is found to be equal to $$RT$$, then

A
The rate of reaction does not depend upon initial concentration

B
The rate constant becomes about $$35$$% of the Arrhenius constant $$A$$

C
The rate constant becomes equal to $$73$$% of the Arrhenius constant $$A$$

D
The rate of the reaction becomes infinite or zero

Solution

$$ k = Ae^{\dfrac{-E_{a}}{RT}} $$

when $$E_{a} = RT$$

$$ k = \dfrac{A}{e} $$

$$\Rightarrow k=0.37 \times A $$

Hence the rate constant becomes approx. 35% of A.
Question 3
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Two reactions: (I) $$A\rightarrow products$$ and (II) $$B\rightarrow products$$, follow first-order kinetics. The rate of reacion-I is doubled when temperature is raised from $$300$$ to $$310K$$. The half-life for this reaction at $$310K$$ is $$30min$$. At the same temperature, $$B$$ decomposes twice as fast as $$A$$. If the energy of activation for the reaction II is half that of reaction I, the rate constant of reaction II at $$300K$$ is

A
$$0.0233{ min }^{ -1 }$$

B
$$0.0327{ min }^{ -1 }$$

C
$$0.0164{ min }^{ -1 }$$

D
$$0.0654{ min }^{ -1 }$$

Solution

For reaction (1)

$$ A \rightarrow Products $$

$$ T_{1} $$ = 300K , $$ T_{2} $$ = 310K

Given :
$$ (\frac{k_{2}(310)}{k_{1}(300)})_{A} = 2 $$

Using Arrhenius equation for reaction (1):

$$ log (\frac{k_{2}(310)}{k_{1}(300)})_{A} = \frac{E_{a}(A)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}}) $$

$$ log 2 = frac{E_{a}(A)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}}) $$

(For reaction 2)

B $$ \rightarrow $$ Products

Since at 310 K, B decomposes twice as fast as A.

$$ \because [K_{2}(310K)]_{A} = 0.0231 min^{-1} $$

$$ \because [K_{2}(310K)]_{B} = 2 \times 0.0231 min^{-1} $$

Using Arrhenius equation for reaction (2)

$$ log (\frac{k_{2}(310)}{k_{1}(300)})_{B} = \frac{E_{a}(B)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}}) $$

Given : $$ (E_{a})_{B} = \frac{1}{2}(E_{a})_{A} $$

Substittute the value of above equation and Comparing equations , we get

$$ (\frac{k_{2}(310)}{k_{1}(300)})_{B} = \frac{1}{2} \times log 2 $$

$$ (\frac{0.0462}{k_{1}(300)})_{B} = 1.414 $$

$$ \therefore {k_{1}(300)} = 0.0327 min^{-1} $$
Question 4
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For a zero-order reaction: $$2{ NH }_{ 3 }(g)\rightarrow { N }_{ 2 }(g)+3{ H }_{ 2 }(g)$$, the rate of reaction is $$0.1atm/s$$. Initially only $${ NH }_{ 3 }(g)$$ was present at $$3atm$$ and the reaction is performed at constant volume and temperature. The total pressure of gases after $$10s$$ from the start of reaction will be

A
$$4atm$$

B
$$5atm$$

C
$$3.5atm$$

D
$$4.5atm$$

Solution
Question 5
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Desorption of a gas from metal surface follows first-order kinetics. The rate constant of desorption can be given by Arrhenius equation. If the desorption of hydrogen on manganese is found to increase $$10$$ times on increasing the temperature from $$600$$ to $$1000 K$$, the activation energy of desorption is :

A
$$6.0\ kcal/mol$$

B
$$6.9\ kcal/mol$$

C
$$3.0\ kcal/mol$$

D
$$57.4\ kcal/mol$$

Solution
Question 6
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Rate of an uncatalyzed first-order reaction at $$T$$ $$K$$ is half of the rate of a catalyzed reaction at $$0.5T$$ $$T$$. If the catalyst lowers the threshold energy by $$20\ kcal$$, what is the activation energy of the uncatalyzed reaction?
$$\left( T=300K,\ln { 2 } =0.7 \right) $$

A
$$39.58\ kcal/mol$$

B
$$19.58\ kcal/mol$$

C
$$40.42\ kcal/mol$$

D
$$20.42\ kcal/mol$$

Solution

Given ,

$$ \frac{k_{2}}{k_{1}} = 2 $$

$$T_1 = 300\ K$$

$$ T_{2} = \frac{T_{1}}{2} $$

As we know,

$$ log(\frac{k_{2}}{k_{1}}) = \frac{E_{2}-E_{1}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$$

Putting values in the above equation, we get;

$$ log(2) = \frac{20-E_{1}}{1.987}(\frac{2}{T_1}-\frac{1}{T_1})$$

$$ 0.7 = \frac{20-E_{1}}{1.987}(\frac{2}{T_1}-\frac{1}{T_1})$$

After solving, we get ;

Activation energy of the uncatalyzed reaction, $$E_{1} = 39.58\ kcal/mol$$

Hence , option A is correct .
Question 7
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Which of the following curves represents a zero order reaction?

A

B

C

D

Solution
Question 8
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Which of the following curves represents a first order reaction?

A

B

C

D

Solution

$$t_{1/2}$$ is independent of $$a$$ for the first order reaction. So, it will be a straight line parallel to concentration axis.
Question 9
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The reaction $$A+B\to C+D;\ \Delta H=25\ kJ/ mole$$ should have an activation energy :

A
$$-25\ kJ/ mole$$

B
$$< 25\ kJ/ mole$$

C
$$> 25\ kJ/ mole$$

D
either answer $$(B)$$ or $$(C)$$ depending upon experiment

Solution

Option C is correct
Question 10
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The plot of concentration of a reactant vs time for a chemical reaction is shown below :
The order of this reaction with respect to the reactant is:

A
$$0$$

B
$$1$$

C
$$2$$

D
not possible to determine from this plot

Solution

The order of this reaction with respect to the reactant is zero-order reaction is reaction in which the rate does not vary with the increase or decrease in the concentration of the reactants.
Therefore the rate of these reactions is always equal to the rate constant ($$K$$) of the specific reactions (since the rate of these reactions is proportional to the 0th power of reactants concentration).

The Differential form of a zero-order reaction can be written as

Rate $$=\dfrac{dA}{dt}=k[A]^{0}=k$$

Hence option A is correct.