Chemical Kinetics Test

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Chemical Kinetics Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

How much faster would a reaction proceed at $$25^0$$C than at $$0^0$$C if the activation energy is $$65$$ kJ?

A
$$2$$ times

B
$$16$$ times

C
$$11$$ times

D
$$6$$ times

Solution

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

$$log \dfrac{{k}_{2}}{{k}_{1}} = \dfrac{65000}{2.303 \times 8.314} (\dfrac{1}{273} - \dfrac{1}{298})$$

$$\dfrac{{k}_{2}}{{k}_{1}}$$ = $$11$$

So, It is 11 times faster.
Question 2
1 / -0

A reaction of first - order completed $$90\% $$ in $$90$$ minutes , hence , it is completed $$50\% $$ in approximately :

A
$$ 50 min $$

B
$$ 54 min $$

C
$$ 27 min $$

D
$$ 62 min $$

Solution

$$k = \dfrac{2.303}{t} log \dfrac {100}{100 -x}$$

$$k = \dfrac{2.303}{90} log \dfrac {100}{100 -90}$$

$$k = 0.0255\space {min}^{-1}$$

$${t}_{1/2} = \dfrac{2.303}{k} log \dfrac {100}{100 -50}$$

$${t}_{1/2} = \dfrac{2.303}{0.0255} log \dfrac {100}{100 -50}$$

$${t}_{1/2} = \dfrac{0.693}{0.0255}$$

$${t}_{1/2} = 27\space min$$

Hence, the correct option is $$C$$
Question 3
1 / -0

Collision frequency of a gas at $$1\ atm$$ pressure is $$Z$$. Its value at $$0.5\ atm$$ will be:

A
$$0.25Z$$

B
$$2Z$$

C
$$0.50Z$$

D
$$Z$$

Solution

$$\rightarrow$$ As the pressure increases collision frequency increases.
$$\rightarrow$$ Vice versa pressure is decreased to half, frequency of collision decreases to half.
Hence option $$C$$ is correct.
Question 4
1 / -0

Collision diameter is least in case of:

A
$${ H }_{ 2 }$$

B
$$He$$

C
$${ CO }_{ 2 }$$

D
$${ N }_{ 2 }$$

Solution

Collision diameter will be least for the low molecular weight.

$$d\space \alpha\space \dfrac{1}{M}$$

In this case, $$Co_2$$ has the highest molecular weight.

So, Option C is correct
Question 5
1 / -0

The half-life of a zero-order reaction is 30 minutes. What is the concentration of the reactant left after 60 minutes?

A
$$25%$$

B
$$50%$$

C
$$6.25%$$

D
$$0$$

Solution

Its a Zero Order Reaction

$${t}_{1/2} = \dfrac{[{A}_{0}]}{2k}$$

$$k = \dfrac{[{A}_{0}]}{60}$$

$$k = \dfrac{[{A}_{0}] - [A]}{t}$$

Substituting the value of k, we get

$${[{A}_{0}]} = {[{A}_{0}]} - [A]$$

$$[A] = 0$$
Question 6
1 / -0

For which order reaction a straight line is obtained along with $$x - axis$$ by plotting a graph between half-life $$\left( {{t_{1/2}}} \right)$$ and initial concentration $$'a'$$.

A
1

B
2

C
3

D
0

Solution

We know that $$t_{\frac{1}{2}}\propto a^{1-n}$$
The graph to be straight line $$t_{\frac{1}{2}}$$ should be independent of $$a$$.
That is $$1-n=0$$, hence $$n=1$$
It is a first order reaction.
Question 7
1 / -0

For the decomposition of $$HI$$ the following logarithmic plot is shown: $$[R=1.98\ cal/mol-K]$$
The activation energy of the reaction is about?

A
$$45600\ cal$$

B
$$13500\ cal$$

C
$$24600\ cal$$

D
$$32300\ cal$$

Solution

$$K=Ae^{-Ea/RT}$$ Slope= $$y/x=\cfrac {1\times 10^3}{0.1}=10$$
$$\Rightarrow \log K=-\cfrac {Ea}{2.303 R}\left(\cfrac {1}{T}\right)+\log A$$
Slope= $$-\cfrac {Ea}{2.303 R}$$
$$\therefore E_a= -slope\times 2.303 R$$
$$=-10\times 2.303 \times 1.98\times 10^3$$
$$= -45.59 \times 10^3$$ cal
Question 8
1 / -0

If $$60\%$$ of a first order reaction was completed in $$60$$ minute, $$50\%$$ of the same reaction would be completed in approximately?

A
$$46$$ minute

B
$$60$$ minute

C
$$40$$ minute

D
$$50$$ minute

Solution

Its a 1st Order reaction.

$$k = \dfrac{2.303}{60} log \dfrac{100}{40}$$

$$k = 0.015\space {min}^{-1}$$

$$t = \dfrac{2.303}{0.015} log \dfrac{100}{50}$$

$$t = 46\space min$$

Hence, the correct option is $$A$$
Question 9
1 / -0

The rate of a first-order reaction is $$0.04\ mol\ L^{-1}s^{-1}$$ at $$10$$ seconds and $$0.03\ mol\ L^{-1}\ s^{-1}$$ at $$20$$ seconds after initiation of the reaction. The half-life period of the reaction is:

A
$$24.1$$ s

B
$$34.1$$ s

C
$$44.1$$s

D
$$54.1$$ s

Solution

For given $$1^{st}$$ order reaction, $$K= \cfrac {1}{t_2-t_1}ln \cfrac {C_1}{C_2}$$

where $$C_1$$ & $$C_2$$ donate concentrations at $$t=10s$$ & $$t=20s$$ respectively.

$$\therefore K= \cfrac {1}{20-10} ln \cfrac {0.04}{0.03}=0.0288$$

$$\therefore t_{1/2}= \cfrac {ln2}{K}= \cfrac {ln2}{0.0288}\approx 24.1s$$

Hence, the correct option is $$A$$
Question 10
1 / -0

The conversion of vinyl allyl ether to pent-4-enol follows a certain kinetics. The following plot is obtained for such a reaction.

A
zero

B
-1

C
1

D
2

Solution

Since the plot of $$\log(a-x)$$ vs $$t$$ is a straight line,

$$\therefore \log(a-x)\propto t$$

This is followed by $$1^{st}$$ order reaction where $$\log (a-x)=\log a_{\circ}-\cfrac{Kt}{2.303}$$

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Chemical Kinetics Test - 46

Result

Chemical Kinetics Test - 46

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Question 1

$$2$$ times

$$16$$ times

$$11$$ times

$$6$$ times

Solution

Question 2

$$ 50 min $$

$$ 54 min $$

$$ 27 min $$

$$ 62 min $$

Solution

Question 3

$$0.25Z$$

$$2Z$$

$$0.50Z$$

$$Z$$

Solution

Question 4

$${ H }_{ 2 }$$

$$He$$

$${ CO }_{ 2 }$$

$${ N }_{ 2 }$$

Solution

Question 5

$$25%$$

$$50%$$

$$6.25%$$

$$0$$

Solution

Question 6

1

2

3

0

Solution

Question 7

$$45600\ cal$$

$$13500\ cal$$

$$24600\ cal$$

$$32300\ cal$$

Solution

Question 8

$$46$$ minute

$$60$$ minute

$$40$$ minute

$$50$$ minute

Solution

Question 9

$$24.1$$ s

$$34.1$$ s

$$44.1$$s

$$54.1$$ s

Solution

Question 10

zero

-1

1

2

Solution

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