Chemical Kinetics Test

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Chemical Kinetics Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

What does it mean when a collision is elastic?

A
No energy is gained or lost.

B
Energy is gained.

C
Energy is lost.

D
The particles can stretch out.

E
The particles slow down.

Solution

(A) , No energy is gained or lost.
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms
Question 2
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Here is another look at the reaction of crystal violet with sodium hydroxide, a first-order reaction (In $$A$$ v time):
What is the significance of the slope?

A
The slope represents the change in concentration over time.

B
The slope represents the inverse of change in concentration over time.

C
The slope represents the negative value of the rate constant.

D
The slope represents the negative rate of product concentration over time.

Solution

By graph,$$\ln { [{ A }_{ t }] } =-Kt+\ln { [{ A }_{ o }] } $$
$$\therefore \ln { \left[ \cfrac { { A }_{ t } }{ { A }_{ o } } \right] } =-Kt\longrightarrow (1)$$
$$ K\longrightarrow $$ Rate constant=slop of graph
$$ [{ A }_{ t }],[{ A }_{ o }]\rightarrow$$ concentrations of reactant at $$t=0,t=5$$ respectively.
It is a first order reaction.
$$ \therefore$$ The slop represents change in concentration over time. (from (1))
Question 3
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In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half the reactant to be consumed:

A
increases two-fold

B
increases four-fold

C
decreases by half

D
does not change

Solution

For a zero order reaction, $$t_{1/2}=\dfrac{[A]_o}{2k}$$

Time increases twice as half-life is directly proportional to initial concentration
Question 4
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If $$50$$% of the reactant is converted into a product in a first order reaction in 25 minutes, how much of it would react in 100 minutes?

A
$$93.75$$%

B
$$87.5$$%

C
$$75$$%

D
$$100$$%

Solution

According to first order reaction kinetics:

$$K=\dfrac{ln\ 2}{25}min^{-1}$$

$$[A]_t=[A]_oe^{-4\ ln2}$$

$$[A]_t=\dfrac{[A]_o}{2^4}=0.0625[A]_o$$

Hence, the percentage will be $$100-6.25=93.75$$%.
Question 5
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A zero-order reaction, $$A \rightarrow$$ Product, with an initial concentration $$ { \left[ A \right] }_{ 0 }$$ has a half-life of $$0.2 s$$. If one starts with the concentration $$2{ \left[ A \right] }_{ 0 }$$, then the half-life is:

A
$$0.1 s$$

B
$$0.4 s$$

C
$$0.2 s$$

D
$$0.8 s$$

Solution

For zero-order reaction, half-life is $$\dfrac{[A]_0}{2k}$$
"or"
$$k=\dfrac{[A]_0}{2\ t_{1/2}}=\dfrac{[A]_0}{0.4}$$

If initial concentration is $$2[A] _0$$ then,

$$t_{1/2}=\dfrac{2[A]_0}{2\times \dfrac{[A]_0}{0.4}}=0.4\ sec$$
Question 6
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Decomposition of $$NH_3$$ on gold surface follows zero order kinetics. If rate constant $$K$$ is $$5\times 10^{-4}M$$ $$s^{-1}$$, rate of formation of $$N_2$$ will be:

A
$$10^{-3}M\ s^{-1}$$

B
$$2.5\times 10^{-4}M \ s^{-1}$$

C
$$5\times 10^{-4}M \ s^{-1}$$

D
zero

Solution

$$N_{2} + 3H_{2} \rightarrow 2NH_{3} $$

Rate $$ = \dfrac{-1}{2} \dfrac{d [NH_{3}]}{dT} = \dfrac{d[N_{2}]}{dT} = \dfrac{1}{3} \dfrac {d[NH_{3}]}{dT} = k = 5 \times 10^{-4} M \ s^{-1}$$

Rate of formation of $$N_{2}$$ $$ = \dfrac{d[N_{2}]}{dT} = k = 5 \times 10^{-4} M \ s^{-1}$$
Question 7
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The rate of a particular reaction triples when temperature changes from $$50^o$$C to $$100^o$$C. What is the activation energy of the reaction (in $$J.mol^{-1}$$)? $$(log 3=0.4771, R=8.314K^{-1}mol^{-1})$$

A
$$24.012\times 10^{-3}$$

B
$$24.012\times 10^3$$

C
$$22.012\times 10^{-3}$$

D
$$22.012\times 10^3$$

Solution

From Arrhenius equation we have,

$$2.303\ log\dfrac{K_2}{K_1}=\dfrac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$

$$2.303\ log\dfrac{3}{1}=\dfrac{E_a}{8.314}(\frac{1}{323}-\frac{1}{373})$$

$$\Rightarrow E_a=22.012\times 10^3\ J$$
Question 8
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$$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$$
What is the ratio of the rate of decomposition of $$N_2O_5$$ to rate of formation of $$NO_2$$?

A
1 : 2

B
2 : 1

C
1 : 4

D
4 : 1
Question 9
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Consider the Arrhenius equation, $$k = Ae^{-E_{a}/(RT)}$$, which of the following would not increase the rate constant of the reaction?

A
A new value of the constant $$A$$ was calculated, this new value being higher than the previous one

B
The temperature of the reaction is increased

C
Experiments have revealed that the previously accepted activation energy of the reaction was calculated incorrectly. The new value is larger.

D
The universal gas constant was redetermined to a smaller value

Solution

$$ k={ A.e }^{ { -E }_{ a }/RT }$$
(i) If the value of $$A$$ increases, the value of $$k$$ increases.
(ii) If the value of $$T$$ increases, $$ { E }_{ a }/RT $$ decreases. Therefore $$ { e }^{ { -E }_{ a }/RT }$$ increases. Therefore, $$k$$ increases.
(iii) If $$ { E }_{ a } $$ increases,$${ e }^{ { -E }_{ a }/RT } $$ decreases. Therefore, $$k$$ decreases.
(iv) If $$R$$ increases, $$ { e }^{ { -E }_{ a }/RT } $$ increases. Therefore, $$k$$ increases.
Question 10
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$$K_{p}$$ of a reaction at $$300\ K$$ is $$6\ atm$$ and $$2\ atm$$ at $$450\ K$$. Which of the following statements is incorrect about this reaction, if $$\triangle n_{g} = 1$$?

A
The reaction is exothermic

B
The rate of backward reaction increases more than that of forward reaction with increase in temperate

C
$$E_{a}$$ for the forward reaction is more than that of backward reaction

D
The difference between heat of reaction at constant pressure and that at constant volume is $$RT$$

Solution

Since the value of the equilibrium constant, $$K_{p}$$ decreases with rise in temperature, it is an exothermic reaction. For an exothermic reaction, the rate of backward reaction increases more than that of forward reaction with the increase in temperature. For such reactions,

$$(E_{a})_{f} < (E_{a})_{b}$$

We know that,
$$\triangle H = \triangle E + \triangle n_{g}RT$$

$$\therefore \triangle H - \triangle E = RT = constant\ [\because \triangle n_{g} = 1(given)]$$.

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Chemical Kinetics Test - 64

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Chemical Kinetics Test - 64

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Question 1

No energy is gained or lost.

Energy is gained.

Energy is lost.

The particles can stretch out.

The particles slow down.

Solution

Question 2

The slope represents the change in concentration over time.

The slope represents the inverse of change in concentration over time.

The slope represents the negative value of the rate constant.

The slope represents the negative rate of product concentration over time.

Solution

Question 3

increases two-fold

increases four-fold

decreases by half

does not change

Solution

Question 4

$$93.75$$%

$$87.5$$%

$$75$$%

$$100$$%

Solution

Question 5

$$0.1 s$$

$$0.4 s$$

$$0.2 s$$

$$0.8 s$$

Solution

Question 6

$$10^{-3}M\ s^{-1}$$

$$2.5\times 10^{-4}M \ s^{-1}$$

$$5\times 10^{-4}M \ s^{-1}$$

zero

Solution

Question 7

$$24.012\times 10^{-3}$$

$$24.012\times 10^3$$

$$22.012\times 10^{-3}$$

$$22.012\times 10^3$$

Solution

Question 8

1 : 2

2 : 1

1 : 4

4 : 1

Question 9

A new value of the constant $$A$$ was calculated, this new value being higher than the previous one

The temperature of the reaction is increased

Experiments have revealed that the previously accepted activation energy of the reaction was calculated incorrectly. The new value is larger.

The universal gas constant was redetermined to a smaller value

Solution

Question 10

The reaction is exothermic

The rate of backward reaction increases more than that of forward reaction with increase in temperate

$$E_{a}$$ for the forward reaction is more than that of backward reaction

The difference between heat of reaction at constant pressure and that at constant volume is $$RT$$

Solution

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