Chemical Kinetics Test

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Chemical Kinetics Test - 67

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Question 1

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Find the values of $$A, B$$ and $$C$$ in the following table for the reaction $$X + Y\rightarrow Z$$. The reaction is of first order w.r.t $$X$$ and zero w.r.t. $$Y$$.

Exp.	$$[X] (mol\ L^{-1})$$	$$[Y] (mol\ L^{-1})$$	Initial rate $$(mol\ L^{-1}s^{-1})$$
$$1.$$	$$0.1$$	$$0.1$$	$$2\times 10^{-2}$$
$$2.$$	$$A$$	$$0.2$$	$$4\times 10^{-2}$$
$$3,$$	$$0.4$$	$$0.4$$	$$B$$
$$4.$$	$$C$$	$$0.2$$	$$2\times 10^{-2}$$

$$A = 0.2\ mol\ L^{-1}, B = 8\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.1\ mol\ L^{-1}$$

$$A = 0.4\ mol\ L^{-1}, B = 4\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.2\ mol\ L^{-1}$$

$$A = 0.2\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$$

$$A = 0.4\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$$

Solution

$$Rate = k[X][Y]^{0}$$

Rate is independent of the conc. of $$Y$$ and it depends only on the conc. of $$X$$ and it is the first-order reaction.

From exp. $$(1), 2\times 10^{-2} = k(0.1) ..... (i)$$

From exp. $$(2), 4\times 10^{-2} = k(A) ....(ii)$$

Dividing (ii) and (i), $$\dfrac {4\times 10^{-2}}{2\times 10^{-2}} = \dfrac {k(A)}{k(0.1)} = \dfrac {A}{0.1}$$

$$\Rightarrow 2\times 0.1 = A$$

$$\Rightarrow A = 0.2\ mol\ L^{-1}$$

From exp. $$(3), B = k(0.4) .... (iii)$$

Dividing (iii) and (i), $$\dfrac {B}{2\times 10^{-2}} = \dfrac {k(0.4)}{k(0.1)} = 4$$

$$\Rightarrow B = 4\times 2\times 10^{-2} = 8\times 10^{-2} mol\ L^{-1} s^{-1}$$

From exp. $$(4), 2\times 10^{-2} = k(C) .....(iv)$$

Dividing (iv) and (i), $$\dfrac {2\times 10^{-2}}{2\times 10^{-2}} = \dfrac {k(C)}{k(0.1)} = \dfrac {C}{0.1}$$

$$\Rightarrow C = 0.1\ mol\ L^{-1}$$.

Question 2
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The energy diagram of a reaction $$P + Q \rightarrow R + S$$ is given. What are $$A$$ and $$B$$ in the graph?

A
$$A \rightarrow$$ activation energy, $$B \rightarrow$$ heat of reaction

B
$$A \rightarrow$$ threshold energy, $$B \rightarrow$$ heat of reaction

C
$$A \rightarrow$$ heat of energy, $$B \rightarrow$$ activation reaction

D
$$A \rightarrow$$ potential energy, $$B \rightarrow$$ energy of reaction

Solution

The heat of reaction is given by:-
$$\triangle H=\quad { Ea }_{ f }\quad -\quad { Ea }_{ b }$$.

Let's suppose $$O$$ is the highest point of the curve then $$RO$$ is forward activation energy and $$PO$$ is backward activation energy and the difference of these two will give the heat of reaction, [Indicated in the figure]

Hence, the correct answer is option $$\text{A}$$.
Question 3
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The activation energy for the reaction-
$$H_{2}O_{2} \rightarrow H_{2}O + \dfrac {1}{2} O_{2}$$
is $$18\ K\ cal/mol$$ at $$300\ K$$. calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Anti log $$(-13.02) = 9.36\times 10^{-14}$$

A
$$9.36\times 10^{-14}$$

B
$$1.2\times 10^{-12}$$

C
$$4.2\times 10^{-30}$$

D
$$5.2\times 10^{-15}$$

Solution

We know that Arrhenius equation for calculation of energy of activation of a reaction is given as:
$$K=A$$ $$e^{-Ea/RT}$$ $$- (i)$$
where, $$K$$= Rate constant
$$A$$= pre exponential factor or frequency factor
$$Ea$$= Activation energy
$$R$$= Ideal gas constant
$$T$$= Temperature in K (Kelvin)

Now, the fraction of moleucle having energy equal to or greater than activation energy is given by:-
$$f= e^{-Ea/RT}$$ $$- (ii)$$
Given, $$Ea=18 K Cal/mol$$ , $$T=300 K$$
$$R=2 CalK^{-1}mol^{-1}$$

Using these values in equation $$(ii)$$
So, $$p=e^{-18\times 1000/2\times 300}$$ $$[\because 1KCal=1000Cal]$$

Taking $$\log$$ on both sides:-
$$\log f=\cfrac {18\times 1000}{2\times 300}=-30$$
$$\Rightarrow f=$$ $$Antilog$$ $$(-30)= 10^{-30}$$
Question 4
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The following data were obtained during the first order thermal composition of $$SO_{2}Cl_{2}$$ at a constant volume.
$$SO_{2}Cl_{2(g)} \rightarrow SO_{2(g)} + Cl_{2(g)}$$
Experiment $$Time/ s^{-1}$$ Total pressure (atm)
$$1$$
$$2$$ $$0$$
$$100$$ $$0.5$$
$$0.6$$
What is the rate of reaction when total pressure is $$0.65\ atm$$?

A
$$0.35\ atm\ s^{-1}$$

B
$$2.235\times 10^{-3}\ atm\ s^{-1}$$

C
$$7.8\times 10^{-4}\ atm\ s^{-1}$$

D
$$1.55\times 10^{-4}\ atm\ s^{-1}$$

Solution

$$\underset {p_o\\ {p_o-p}}{SOCl_2} \rightarrow \underset {0\\p}{SO_2} + \underset {0\ at\ t=0\\p\ at\ t=t}{Cl_2}$$

Pressure at time $$t, P_{t} = p_{0} - p + p + p = p_{0} + p$$

Pressure of reactants at time $$t, p_{0} - p = 2p_{0} - p_{t}\propto R$$

$$k = \dfrac {2.303}{t} \log \dfrac {p_{0}}{2p_{0} - P_{t}}$$

$$= \dfrac {2.303}{100}\log \dfrac {0.5}{2\times 0.5 - 0.6} = \dfrac {2.303}{100}\log 1.25$$

$$= 2.2318\times 10^{-3} s^{-1}$$

Pressure of $$SO_{2}Cl_{2}$$ at time $$t(p_{SO_{2}Cl_{2}})$$

$$= 2p_{0} - P_{t} = 2\times 0.50 - 0.65\ atm = 0.35\ atm$$

Rate at that time $$= k\times p_{SO_{2}Cl_{2}}$$

$$= (2.2318\times 10^{-3})\times (0.35) = 7.8\times 10^{-4} atm\ s^{-1}$$.
Question 5
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How much faster would a reaction proceed at 298 K than at 273 K, if the activation energy is $$65 kJ mol^{-1}$$?

A
22 times

B
11 times

C
33 times

D
5.5 times

Solution

$$ \because k = Ae^{-Ea/RT}$$
$$ \implies \dfrac{k_{25}}{k_0} = \dfrac{A_{25} e^{65000/8.314 \times 298}}{A_{0} e^{65000/8.314 \times 275}}$$
$$ \implies \dfrac{e^{-26.24}}{e^{-28.64}}$$
$$ \implies e^{2.40} = 11 $$
Thus, the reaction will proceed 11 times faster.
Question 6
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The reaction at hydrogen and iodine monochloride is represented by the equation is
$$H_{2} (g) + 2ICl (g) \rightarrow 2HCl (g) + I_{2} (g)$$
This reaction is first order in $$H_{2}(g)$$ and also first - order in ICI(g). which of these proposed mechanism can be consistent with the given information about this reaction ?
Mechanism I : $$H_{2} (g) + 2ICl (g) \rightarrow 2HCl (g) + I_{2} (g)$$
Mechanism II : $$H_{2} (g) + ICl (g) \overset{slow}{\rightarrow} HCl (g) + HI (g)$$
$$HI (g) + ICl (g) \overset{fast}{\rightarrow} HCl (g) + I_{2} (g)$$ :

A
I only

B
II only

C
Both I and II

D
Neither I nor II

Solution

According to question the given reaction is first order in $$H_2(g)$$ and also first order in $$ICl(g)$$ so, the slowest (rate determining) step in the reaction should involve $$H_2(g)$$ & $$ICl(g)$$ one mole each.
The mechanism $$II$$ is seeming consistent with the conditions stated as:-
$$H_2(g)+ICl(g)\longrightarrow HCl(g)+HI(g)$$ (slow)
$$Rate=K[H_2][ICl]$$
& $$HI(g)+ICl(g)\longrightarrow HCl(g)+I_2(g)$$ (fast)
Question 7
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The activation energy for a reaction is $$9.0 k cal/mol$$. The increase in the rate constant when its temperature is increased from $$298 K$$ to $$308 K$$ is:

A
$$63$$%

B
$$50$$%

C
$$100$$%

D
$$10$$%

Solution

$$2.303 \log \dfrac{K_2}{K_1} = \dfrac{E_a}{R} \left[\dfrac{T_2 - T_1}{T_1T_2}\right]$$

$$\log \dfrac{K_2}{K_1} = \dfrac{9.0 \times 10^3}{2.303 \times 2} \left[\dfrac{308 - 298}{308 \times 298}\right]$$

$$\dfrac{K_2}{K_1} = 1.63; K_2 = 1.63 \,K_1; \dfrac{1.63 K_1 - K_1}{K_1} \times 100 = 63.0 \%$$
Question 8
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For a reaction $$A_{2} + B_{2}\rightleftharpoons 2AB$$ the figure shows the path of the reaction in absence and presence of a catalyst. What will be the energy of activation for forward $$(E_{f})$$ and backward $$(E_{b})$$ reaction in presence of a catalyst and $$\triangle H$$ for the reaction? the dotted curve is the path of reaction in presence of a catalyst.

A
$$E_{f} = 60\ kJ/mol, E_{b} = 70\ kJ/ mol, \triangle H = 20\ kJ/ mol$$

B
$$E_{f} = 20\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = 50\ kJ/ mol$$

C
$$E_{f} = 70\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = 10\ kJ/ mol$$

D
$$E_{f} = 10\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = -10\ kJ/ mol$$

Solution

$$E_{f} = 70 - 60= 10\ kJ/mol$$

$$E_{b} = 70 - 50 = 20\ kJ/ mol$$

$$\triangle H = 50 - 60 = -10\ kJ/ mol$$

Option D is correct
Question 9
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The fraction of collisions that posses the energy $$E_a$$ is given by:

A
$$f = e^{\frac{-Ea}{RT}}$$

B
$$f = e^{\frac{Ea}{RT}}$$

C
$$f = e^{- Ea.RT}$$

D
$$f = e^{Ea.RT}$$

Solution

$$k=Pze^{\cfrac {-Ea}{RT}}$$
$$\cfrac kA$$ or $$\cfrac k{Pz}= $$ Fraction of collision possessing $$E_a$$
$$\therefore$$ Fraction of collision possessing energy $$E_a= \cfrac kA$$
$$=\cfrac {Ae^{-E_a/RT}}{A}$$
$$f=e^{-E_a/RT}$$
Question 10
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For an exothermic reaction $$A \rightarrow B$$, the activation energy is $$65\ kJ\ mol^{-1}$$ and heat of reaction $$-42\ kJ\ mol^{-1}$$. The activation energy for the reaction $$B \rightarrow A$$ would be :-

A
$$23\ kJ\ mol^{-1}$$

B
$$107\ kJ\ mol^{-1}$$

C
$$65\ kJ\ mol^{-1}$$

D
$$42\ kJ\ mol^{-1}$$

Solution

$$A \rightarrow B$$
$$(E_a)_1=65 \ kJ/mol$$
$$\triangle H=-42 \ kJ/mol$$
Now, For $$B \rightarrow A$$
$$(E_a)_2=(E_a)_1- \triangle H$$
$$=65 -(-42)=107 \ kJ/mol$$

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Experiment	$$Time/ s^{-1}$$	Total pressure (atm)
$$1$$ $$2$$	$$0$$ $$100$$	$$0.5$$ $$0.6$$

Chemical Kinetics Test - 67

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Chemical Kinetics Test - 67

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Question 1

$$A = 0.2\ mol\ L^{-1}, B = 8\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.1\ mol\ L^{-1}$$

$$A = 0.4\ mol\ L^{-1}, B = 4\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.2\ mol\ L^{-1}$$

$$A = 0.2\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$$

$$A = 0.4\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$$

Solution

Question 2

$$A \rightarrow$$ activation energy, $$B \rightarrow$$ heat of reaction

$$A \rightarrow$$ threshold energy, $$B \rightarrow$$ heat of reaction

$$A \rightarrow$$ heat of energy, $$B \rightarrow$$ activation reaction

$$A \rightarrow$$ potential energy, $$B \rightarrow$$ energy of reaction

Solution

Question 3

$$9.36\times 10^{-14}$$

$$1.2\times 10^{-12}$$

$$4.2\times 10^{-30}$$

$$5.2\times 10^{-15}$$

Solution

Question 4

$$0.35\ atm\ s^{-1}$$

$$2.235\times 10^{-3}\ atm\ s^{-1}$$

$$7.8\times 10^{-4}\ atm\ s^{-1}$$

$$1.55\times 10^{-4}\ atm\ s^{-1}$$

Solution

Question 5

22 times

11 times

33 times

5.5 times

Solution

Question 6

I only

II only

Both I and II

Neither I nor II

Solution

Question 7

$$63$$%

$$50$$%

$$100$$%

$$10$$%

Solution

Question 8

$$E_{f} = 60\ kJ/mol, E_{b} = 70\ kJ/ mol, \triangle H = 20\ kJ/ mol$$

$$E_{f} = 20\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = 50\ kJ/ mol$$

$$E_{f} = 70\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = 10\ kJ/ mol$$

$$E_{f} = 10\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = -10\ kJ/ mol$$

Solution

Question 9

$$f = e^{\frac{-Ea}{RT}}$$

$$f = e^{\frac{Ea}{RT}}$$

$$f = e^{- Ea.RT}$$

$$f = e^{Ea.RT}$$

Solution

Question 10

$$23\ kJ\ mol^{-1}$$

$$107\ kJ\ mol^{-1}$$

$$65\ kJ\ mol^{-1}$$

$$42\ kJ\ mol^{-1}$$

Solution

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