Chemical Kinetics Test

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Chemical Kinetics Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

In a first order reaction the amount of reactant decayed in three half lives (let a be is initial amount) would be:

A
7a/8

B
a/8

C
a/6

D
5a/6

Solution

The amount of reactant left after $$n$$ half lives for a first order reaction is $$(1/2)n$$
times the Initial concentration.
Here $$n=3$$ . So Amount of Reactant left is $$\dfrac{1}{8}$$
So Amount of reactant decayed is $$a-\dfrac{1}{8}=7a/8$$
Option A is correct.
Question 2
1 / -0

The rate of a reaction at 200 K 100 times less than the rate at 400 K.The activation energy of the reaction is

A
184 R

B
1840 R

C
460 R

D
4600 R

Solution
Question 3
1 / -0

Rate constant of a first order reaction is $$1.15\times 10^{-5}$$. The percentage of initial concentration that remained after 1 hour is?

A
$$89.15\%$$

B
$$95.96\%$$

C
$$55.86\%$$

D
$$83.45\%$$

Solution

Given: $$I^{st}$$ order reaction,
$$k=1.15\times 10^{-5}$$

$$t=1\ hour=60\times 60=3600\ seconds$$
We have :

$$k=\dfrac{2.303}{t}\dfrac{Initial\ conc}{final\ Conc}$$

On putting all given values:

$$1.15\times 10^{-5}=\dfrac{2.303}{3600}\dfrac{I_c}{F_c}$$

$$I_c$$= initial concenmtration, $$f_c$$= final concenmtration
On solving we will get $$F_c=95.96%$$%
Option B is correct.
Question 4
1 / -0

For a first order process $$A\rightarrow B$$ rate constant $${k}_{1}=0.693\ {min}^{-1}$$ and another first order process $$C\rightarrow D$$, $${K}_{2}=x$$ $${min}^{-1}$$. If $$99.9$$% of $$C\rightarrow D$$ requires time same as $$50$$% of reaction $$A\rightarrow B$$, value of $$x$$? (in $${min}^{-1}$$)

A
$$0.0693$$

B
$$6.93$$

C
$$23.03$$

D
$$13.86$$

Solution

The solution is as follows:

1. Time required for $$50$$% completion for reaction $$1$$ is:
$$t_{1/2}=\dfrac{0.693}{k}$$
Given $$k=0.639\ min^{-1}$$

So $$t_{1/2}=\dfrac{0.693}{0.693}=1\ min$$

2. For reaction $$2$$, it takes $$1$$ min to get $$99.99$$% of reaction to be completed. Thus, it takes $$0.5$$ min to get $$50$$% reaction.

Thus,
$$t_{1/2}=\dfrac{0.693}{K_2}$$

$$K_2=\dfrac{0.693}{t_{1/2}}$$

$$K_2=\dfrac{0.693}{0.5}$$

Option D is correct.
$$K_2=13.86\ min^{-1}$$
Question 5
1 / -0

A reaction $$A+B\leftrightarrow C+D$$ follows the mechanism:
$$A+B\leftrightarrow AB$$
$$AB+C\leftrightarrow D$$
In which first step remains essentially in equilibrium. If $$\Delta H$$ is the enthalpy change for the first reaction the activation energy for the second reaction, the activation energy of the overall reaction will be given?

A
$$E_{0}$$

B
$$E_{0}-\Delta H$$

C
$$E_{0}+\Delta H$$

D
$$E_{0}+2\Delta H$$
Question 6
1 / -0

Raw milk sours in 4 hours at $$27^\circ C,$$ but in 40 hours in refrigerator at $$7^\circ C.$$ What is activation energy for souring of milk:

A
402.1 J/mol

B
30.2 J/mol

C
8.04 KJ/mol

D
80.4 KJ/mol

Solution

$$K\alpha\dfrac{1}{t}$$

$$Log\dfrac{K_2}{K_1}=log\dfrac{t_1}{t_2}=\dfrac{E_a}{R}(\dfrac{T_2-T_1}{T_I\times T_2})$$

$$Log\dfrac{40}{4}=\dfrac{E_a}{8.314}(\dfrac{20}{300\times 280})$$

So solving $$E_a=30.2J/mol$$
Question 7
1 / -0

For a first order reaction, the plot of $$'t'$$ against $$log C$$ gives a straight line with slope equal to:

A
$$(k/2.2303)$$

B
$$(-k/2.303)$$

C
$$(In k/2.303)$$

D
$$-k$$

Solution

For a first-order reaction,
$$log(a-x)=-\dfrac{kt}{2.303}+loga$$

It is similar to straight-line equation $$y=mx+c$$

where intercept is equal to $$loga$$ and slope equal to $$-\dfrac{k}{2.303}$$

Option B is the correct answer.
Question 8
1 / -0

For the reaction : $$ NH_2COONH_4(s) \leftrightharpoons 2NH_3(g) +CO_2(g) , K_p = 3.2 \times 10^{-5}atm^3 $$
the total pressure of the gaseous products when sufficient amount of reactant is allowed to achieve equilibrium , is:

A
$$0.02$$ atm

B
$$0.04$$ atm

C
$$0.06$$ atm

D
$$0.095$$ atm

Solution
Question 9
1 / -0

A zero-order reaction is one

A
in which reactants do not react

B
in which one of the reactants is in large excess

C
whose rate does not change with time

D
whose rate increases with time

Solution

In a zero-order reaction, the rate is independent of the concentration of the reactants. The rate of change of concentration would be constant.

Therefore, $$Rate=k=constant$$

Hence, the correct option is (C).
Question 10
1 / -0

The temperature coefficient of a reaction is $$2$$. When the temperature is increased from $$30^{\circ}C$$ to $$90^{\circ}C$$, the rate of reaction is increased by

A
$$150$$ times

B
$$410$$ times

C
$$72$$ times

D
$$64$$ times

Solution

For every $$10^{\circ}C$$ rise in temperature the rate of reaction doubles.

$$\therefore$$ From $$30^{\circ}C$$ to $$90^{\circ}C$$ there 6 times $$10^oC$$ rise in temperature.

So, rate increase by $$2^6=64$$ times.

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 53

Result

Chemical Kinetics Test - 53

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SHARING IS CARING

Question 1

7a/8

a/8

a/6

5a/6

Solution

Question 2

184 R

1840 R

460 R

4600 R

Solution

Question 3

$$89.15\%$$

$$95.96\%$$

$$55.86\%$$

$$83.45\%$$

Solution

Question 4

$$0.0693$$

$$6.93$$

$$23.03$$

$$13.86$$

Solution

Question 5

$$E_{0}$$

$$E_{0}-\Delta H$$

$$E_{0}+\Delta H$$

$$E_{0}+2\Delta H$$

Question 6

402.1 J/mol

30.2 J/mol

8.04 KJ/mol

80.4 KJ/mol

Solution

Question 7

$$(k/2.2303)$$

$$(-k/2.303)$$

$$(In k/2.303)$$

$$-k$$

Solution

Question 8

$$0.02$$ atm

$$0.04$$ atm

$$0.06$$ atm

$$0.095$$ atm

Solution

Question 9

in which reactants do not react

in which one of the reactants is in large excess

whose rate does not change with time

whose rate increases with time

Solution

Question 10

$$150$$ times

$$410$$ times

$$72$$ times

$$64$$ times

Solution

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