Chemical Kinetics Test

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Chemical Kinetics Test - 80

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Weekly Quiz Competition

Question 1
1 / -0

Under what condition the order of the reaction,
$$2HI(g)\rightarrow H_2(g)+I_2(g)$$, is zero.

A
At high temperature

B
At high partial pressure of HI

C
At low partial pressure of HI

D
At high partial pressure of $$H_2$$

E
At high partial pressure of $$I_2$$

Solution

The reaction,

$$2HI(g)\overset{\Delta}{\longrightarrow}H_2(g)+I_2(g)$$

is of zero-order in which HI is present at high partial pressure.
Question 2
1 / -0

$$50\%$$ of a zero order reaction completes in $$10$$ minutes. $$100\%$$ of the same reaction shall complete in:

A
$$5\ min$$

B
$$10\ min$$

C
$$20\ min$$

D
$$\infty $$ time

Solution

For zeroth order reaction:

$$At=Ao-Kt$$;

After $$50\%$$ completion, At $$=\dfrac{Ao}{2}$$

$$A/q, \dfrac{Ao}{2}=Ao-K\times 10\Rightarrow K=\dfrac{Ao}{20}$$

Now for $$100\%$$ completion $$At=0$$

$$0=Ao-\dfrac{Ao}{20}\times t\Rightarrow t=20\min$$
Question 3
1 / -0

For a zero order reaction with the initial reactant concentration a, the time for completion of the reaction is

A
$$k/a$$

B
$$a/k$$

C
$$2k/a$$

D
$$a/2k$$

Solution

For zero order reaction .
dQ/dt=k
Q= concentration of reactant at time t .
And k = rate. Constant.

On integrating above equation
Q - Qo = k ( t-to)
Qo = initial concentration and to = start time
So at t= 0; Q= a

At end of the reaction, reactant Concentration=0
a = kt
So, t = a/ k
Question 4
1 / -0

In which of the following reactions of the following orders the molecularity and order can never be same?

A
Zero order

B
First order

C
Second order

D
Third order

Solution

Option A is correct because molecularity can never be zero.

So, for zero-order reaction, molecularity and order can never be same.
Question 5
1 / -0

A plot of reactant concentration versus time for a reaction is a straight line with a negative slope giving the rate constant, and the intercept, giving the initial concentration of the reactant. The order of the reactant is:

A
$$0$$

B
$$1$$

C
$$2$$

D
None of these

Solution
Question 6
1 / -0

For a reaction of the order of $$0.5$$, when the concentration of the reactant is doubled the rate

A
doubles

B
increases four times

C
decrease four times

D
increases $$\sqrt{2}$$ times

Solution

The expression for the rate of the reaction is $$ r=k[A]^n$$

now, $$ r=k[A]^{1/2}......(1)$$

When the concentration of A is doubled, then, rate:

$$r′=k[2A]^{1/2}......(2)$$

Dividing (2) by (1), we get

$$\dfrac{k[2A]^{1/2}}{k[A]^{1/2}}=\sqrt2$$
Question 7
1 / -0

The number of collisions depend upon

A
Pressure

B
Concentration

C
Temperature

D
All the above

Solution

Number of collision depend upon pressure, concentration and temperature.

Option D is correct.
Question 8
1 / -0

The time for the half-life period of a certain reaction $$A\to$$ products is $$1$$ hour. When the initial concentration of the reactant, $$A$$ is $$2.0\ mol\ L^{-1}$$. How much time does it take initial concentration to make from $$0.50$$ to $$0.25\ mol\ L^{-1}$$ if it is a zero-order reaction?

A
$$4\ h$$

B
$$0.5\ h$$

C
$$0.25\ h$$

D
$$1\ h$$

Solution

The initial concentration of A is $$\mathrm{2 \ M}$$.

If the reaction would be first order reaction, then

$$\mathrm{k = \cfrac{a_o - 0.5a_o}{t} = 1}$$

Time taken for the reactant concentration to decrease from 0.5 to 0.25;

$$\mathrm{t = \cfrac{0.5 - 0.25}{k} = \cfrac{0.25}{1}}$$

So, the time taken for reactant concentration to decrease from 0.5 M to 0.25 M is $$0.25 \ h$$.
Hence, Option "C" is the correct answer.
Question 9
1 / -0

A graph between $$t_{1/2}$$ and concentration for $$n^{th}-order$$ reaction is a straight line. The reaction of this nature is completed $$50\%$$ in $$10$$ minutes when concentration is $$2\ mol. L^{−1}.$$ This is decomposed $$50\%$$ in t minutes at $$4$$ $$mol. L^{−1}$$. Then $$n$$ and $$t$$ are respectively:

A
0, 20 min

B
1, 10 min

C
1, 20 min

D
0, 5 min

Solution

It can be observed from the given graph that $$t_{1/2}$$ is constant and is independent of the concentration. We know that,

$$t_{1/2}∝(1/a_o)^{n−1}$$ ---- 1

where,
$$a_o$$ is the initial concentration of the reactant
n is the order of the reaction.

Now for first order reaction $$n=1$$ and $$t_{1/2}$$ will be independent of $$a_o$$ (i.e. concentration) as depicted in the graph.
Hence, the reaction is of $$\text{first order.}$$

As, $$t_{1/2}$$ is independent of initial concentration, it will be the same whether the concentration is $$2\ mol /L$$ or $$4\ mol /L$$
So, $$n=1$$ and $$t=10$$ minutes (as already given in the question)

Option B is correct.
Question 10
1 / -0

For a first order reaction $$A\xrightarrow [ ]{ k } B$$, the degree of dissociation is equal to

A
$$e^{-kt}$$

B
$$1-e^{-kt}$$

C
$$e^{kt}$$

D
$$1+e^{-kt}$$

Solution

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 80

Result

Chemical Kinetics Test - 80

Score

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Rank

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SHARING IS CARING

Question 1

At high temperature

At high partial pressure of HI

At low partial pressure of HI

At high partial pressure of $$H_2$$

At high partial pressure of $$I_2$$

Solution

Question 2

$$5\ min$$

$$10\ min$$

$$20\ min$$

$$\infty $$ time

Solution

Question 3

$$k/a$$

$$a/k$$

$$2k/a$$

$$a/2k$$

Solution

Question 4

Zero order

First order

Second order

Third order

Solution

Question 5

$$0$$

$$1$$

$$2$$

None of these

Solution

Question 6

doubles

increases four times

decrease four times

increases $$\sqrt{2}$$ times

Solution

Question 7

Pressure

Concentration

Temperature

All the above

Solution

Question 8

$$4\ h$$

$$0.5\ h$$

$$0.25\ h$$

$$1\ h$$

Solution

Question 9

0, 20 min

1, 10 min

1, 20 min

0, 5 min

Solution

Question 10

$$e^{-kt}$$

$$1-e^{-kt}$$

$$e^{kt}$$

$$1+e^{-kt}$$

Solution

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