Chemical Kinetics Test

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Chemical Kinetics Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The energy of activation for a first order reaction is $$104 kJ mol^{-1}$$. The rate constant at $$25^oC$$ is $$3.7\times 10^{-5} s^{-1}$$. What is the rate constant at $$30^oC$$ ?

A
$$5.2\times 10^{-4}s^{-1}$$

B
$$7.4 \times 10^{-5} s^{-1}$$

C
$$7.72\times 10^{-5}s^{-1}$$

D
None of these

Solution

$$log \dfrac{{k}_{2}}{{k}_{1}}$$ = $$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$$

Substituting the values,

$$log \dfrac{{k}_{2}}{3.7 \times {10}^{-5}}$$ = $$\dfrac{104 \times {10}^{3}}{8.314 \times 2.303} (\dfrac{1}{298} - \dfrac{1}{303})$$

$${k}_{2} = 7.4 \times {10}^{-5}\space {s}^{-1}$$
Question 2
1 / -0

Find out the percentage of the reactant molecules crossing over the activation energy barrier at 325K, given that $$\triangle { H }_{ 325 }=0.12kcal,\quad { E }_{ a(b) }=+0.02kcal$$.

A
0.80%

B
80.63%

C
20%

D
None of these

Solution

$$\Delta H={ E }_{ a(f) }-{ E }_{ a(b) }$$
(activation energy of forward reaction-action energy of backward reaction)
$$ 0.02+0.12={ E }_{ a(f) }$$
$$ { E }_{ a(f) }= 0.14 kcal/mole.k= 140 cal/mol.K$$
Only those molecules will cross the activation energy barrier which possess energy greater than $${ E }_{ a }.$$
Formula:
Fraction of molecules with energy greater than $${ E }_{ a }=x={ e }^{ -\left( \cfrac { { E }_{ a } }{ RT } \right) }\\ x={ e }^{ -\left( \cfrac { 140 }{ 2\times 325 } \right) }$$ Where $$R=1.98cal/molek,\quad R\approx 2cal/mol.k\\ x={ e }^{ -0.2154 }\\ \ln { x } =0.2154\\ \ln { x } =\cfrac { -0.2154 }{ 2.303 } =0.0935\\ x=antilog(0.0935)\\ x=0.8063\\ Percentage=80.63\%$$
Question 3
1 / -0

Figure depicts the charge in conc. of species $$x$$ & $$y$$ for reaction $$2x \rightarrow y$$, as a function of time the point of inter section of two curves represents?

A
$$t_{1/2}$$

B
$$t_{1/3}$$

C
$$t_{1/4}$$

D
$$t_{2/3}$$

Solution

The point of intersection of two curves represents the time at which half of the reactants are converted to products i.e Half Life ($${t}_{1/2}$$)
Question 4
1 / -0

In a first-order reaction the concentration of the reactant is decreased from $$1.0{\text{ M to 0}}{\text{.25 M }}$$ in 20 min.The rate constant of the reaction would be:

A
$$10{\text{ mi}}{{\text{n}}^{ - 1}}$$

B
$$6.931{\text{ mi}}{{\text{n}}^{ - 1}}$$

C
$$0.6931{\text{ mi}}{{\text{n}}^{ - 1}}$$

D
$$0.06931{\text{ mi}}{{\text{n}}^{ - 1}}$$

Solution

Its First Order Reaction

$$k = \dfrac{2.303}{t} log\space\dfrac {[A]}{[A - x]}$$

Substituting the values,

$$k = \dfrac{2.303}{20} log\space\dfrac {[1]}{[0.25]}$$
$$k = 0.06931\space {min}^{-1}$$
Question 5
1 / -0

For a reversible reaction : $${ N }_{ 2 }+{ O }_{ 2 }\rightleftharpoons \quad 2NO$$
Activation energy of the backward reaction is lower than that of forward reaction. The slope of k verse 1/T graph will be:

A
zero

B
$$-\frac { H }{ 2.303\quad R } $$

C
$$\frac { H }{ 2.303\quad R } $$

D
$$-\frac { \Delta H }{ \quad R }$$

Solution

For reactions whose forward direction has less activation energy, the graph has -ve slope, indicating, a faster reaction in the forward direction.
It is given in the question that the backward reaction has a lower $${ E }_{ A }$$ than the forward reaction. this means graph has a positive slope.
Slope of graph$$=\cfrac { \Delta H }{ 2.303R } $$
Hence answer is [C].
Question 6
1 / -0

A first order reaction is half-completed in $$45$$ minutes. How long does it need for $$99.9\%$$ of the reaction to be completed?

A
$$20$$ hours

B
$$10$$ hours

C
$$7\ \dfrac{3}{10}$$ hours

D
$$5$$ hours

Solution

$${ t }_{ 1/2 }=45$$ minutes

For 1st order reaction $$ { t }_{ 1/2 }=\cfrac { 0.693 }{ k } \\ \therefore k=\cfrac { 0.693 }{ 45 } { min }^{ -1 }=0.0154{ min }^{ -1 }$$

Again for 1st order reaction we have

$$ k=\cfrac { 2.303 }{ t } \log { \left[ \cfrac { 100 }{ 100-99.9 } \right] } $$

(considering original concentration to be 100)

$$ t=\cfrac { 2.303 }{ k } \log { \left[ \cfrac { 100 }{ 0.1 } \right] } $$

$$ t=\cfrac { 2.303 }{ 0.0154 } \log { 1000 } $$

$$ t=\cfrac { 2.303 }{ 0.0154 } \times 3=448.636\quad minutes$$

$$ t=7 \dfrac{3}{10}\quad hours.$$

Hence, the correct option is $$C$$
Question 7
1 / -0

For a first order reaction,

$$(A)$$ $$\rightarrow$$ products,

the concentration of $$A$$ changes from $$0.1$$M to $$0.025$$M in $$40$$ minutes. The rate of reaction when the concentration of $$A$$ is $$0.01$$ M is:

A
$$1.73\times 10^{-4}$$ mol $$dm^{-3} min^{-1}$$

B
$$1.73\times 10^{-5}$$ mol $$dm^{-3} min^{-1}$$

C
$$3.47\times 10^{-4}$$ mol $$dm^{-3} min^{-1}$$

D
$$3.47\times 10^{-5}$$ mol $$dm^{-3} min^{-1}$$

Solution

For $$1^{st}$$ order reaction,

rate constant, $$R=\cfrac {2.303}{t}\log\cfrac {[A]}{[A]_t}$$

$$\therefore R=\cfrac {2.303}{40}\log \cfrac {0.1}{0.025}=\cfrac {0.693}{20}$$

For $$1^{st}$$ order reaction,

rate=$$K[A]=\cfrac {0.693}{20}\times 0.01$$
$$=3.47 \times 10^{-4} mol\ dm^{-3} min^{-1}$$

Hence, the correct option is $$C$$
Question 8
1 / -0

The reaction : $$X \rightarrow$$ product, follows first-order kinetics in $$20$$ minutes, the concentration of $$X$$ changes from $$0.1 M$$ to $$0.05\ M$$ then rate of reaction when concentration of $$X$$ is $$0.02\ mol/ L$$ is:

A
$$1.73 \times 10^{-4} M/min$$

B
$$3.47 \times 10^{-5} M/min$$

C
$$6.94 \times 10^{-4} M/min$$

D
$$1.73 \times 10^{-5} M/min$$

Solution

By first order kinetic rate constant,

$$ k = \cfrac{2.303}{t} \log(\cfrac{a}{a-x})$$

$$a = 0.1 \space M$$

$$ (a-x) = 0.05 \space M$$

$$ t = 20 \space min$$
$$ \implies k = \cfrac{2.303}{20} \log(\cfrac{0.1}{0.05})$$

$$ \implies k = 0.0347 \space min^{-1}$$

$$ Rate= \cfrac{dx}{dt} = k[A]^1$$

$$\implies 0.0347 \times 0.02$$

$$ \implies 6.94 \times 10^{-4} \times M \space min^{-1}$$
Question 9
1 / -0

For zero order reactions, the linear plot was obtained for $$[A]$$ vs t. The slope of the line is equal to:

A
$$k_{0}$$

B
$$-k_{0}$$

C
$$\dfrac {0.693}{K_{o}}$$

D
$$-\dfrac {K_{o}}{2.303}$$

Solution

For a Zero Order Reaction,
$$-\dfrac{d[A]}{dt} = k$$
Integrating on both sides and finding the value of 'c'(Constant of proportionality), we get
('c' can be found out by substituting [A] = $$[{A}_{0}]$$ at t = 0)
$$[A] = -{k}_{0}t + [{A}_{0}]$$
Where, $$-{k}_{0}$$ is rate constant, $$[{A}_{0}]$$ is Initial Concentration.

So, the Slope of the line is $$-{k}_{0}$$
Question 10
1 / -0

An endothermic reaction $$A\rightarrow B$$ has an activation energy $$15kcal/mole$$ and the heat of reaction is $$5kcal/mole$$. The activation energy of reaction $$B\rightarrow A$$ is:

A
$$20kcal/mole$$

B
$$15kcal/mole$$

C
$$10kcal/mole$$

D
zero

Solution

Activation energy=$$ 15 kcal/mole$$
Heat of reaction=$$5 kcal/mole$$
Total $${ E }_{ a }=20kcal/mol$$
For $$A\rightarrow B$$
For any reaction the activation energy in the forward direction is numerically equal to activation energy of reaction in backward direction.
Therefore $${ E }_{ a }$$ for $$B\rightarrow A$$ is 20 kcal/mole and option [A] is correct answer.

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 43

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Chemical Kinetics Test - 43

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SHARING IS CARING

Question 1

$$5.2\times 10^{-4}s^{-1}$$

$$7.4 \times 10^{-5} s^{-1}$$

$$7.72\times 10^{-5}s^{-1}$$

None of these

Solution

Question 2

0.80%

80.63%

20%

None of these

Solution

Question 3

$$t_{1/2}$$

$$t_{1/3}$$

$$t_{1/4}$$

$$t_{2/3}$$

Solution

Question 4

$$10{\text{ mi}}{{\text{n}}^{ - 1}}$$

$$6.931{\text{ mi}}{{\text{n}}^{ - 1}}$$

$$0.6931{\text{ mi}}{{\text{n}}^{ - 1}}$$

$$0.06931{\text{ mi}}{{\text{n}}^{ - 1}}$$

Solution

Question 5

zero

$$-\frac { H }{ 2.303\quad R } $$

$$\frac { H }{ 2.303\quad R } $$

$$-\frac { \Delta H }{ \quad R }$$

Solution

Question 6

$$20$$ hours

$$10$$ hours

$$7\ \dfrac{3}{10}$$ hours

$$5$$ hours

Solution

Question 7

$$1.73\times 10^{-4}$$ mol $$dm^{-3} min^{-1}$$

$$1.73\times 10^{-5}$$ mol $$dm^{-3} min^{-1}$$

$$3.47\times 10^{-4}$$ mol $$dm^{-3} min^{-1}$$

$$3.47\times 10^{-5}$$ mol $$dm^{-3} min^{-1}$$

Solution

Question 8

$$1.73 \times 10^{-4} M/min$$

$$3.47 \times 10^{-5} M/min$$

$$6.94 \times 10^{-4} M/min$$

$$1.73 \times 10^{-5} M/min$$

Solution

Question 9

$$k_{0}$$

$$-k_{0}$$

$$\dfrac {0.693}{K_{o}}$$

$$-\dfrac {K_{o}}{2.303}$$

Solution

Question 10

$$20kcal/mole$$

$$15kcal/mole$$

$$10kcal/mole$$

zero

Solution

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