Chemical Kinetics Test

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Chemical Kinetics Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

Among the following, the maximum covalent character is shown by the compound

A
$$MgCl_{2}$$

B
$$FeCl_{2}$$

C
$$SnCl_{2}$$

D
$$AlCl_{3}$$

Solution

According to Fajans rule, cation with greater charge and smaller size favours covalency
Question 2
1 / -0

Nitric oxide, NO, and bromine vapour react together according to the following equation.
$$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$$
$$\Delta H^o=-23$$kJ$$mol^{-1}$$
The reaction has an activation energy of $$+5.4$$ kJ $$mol^{-1}$$.
What is the correct reaction pathway diagram for this reaction?

A

B

C

D

Solution

$${2 NO}_{\left( g \right)} + {{Br}_{2}}_{\left( g \right)} \longrightarrow {2 NOBr}_{\left( g \right)}$$
$$\Delta{{H}^{°}} = \Delta{{H}_{P}} - \Delta{{H}_{R}}$$
$$\Delta{{H}^{°}} = - 23 \; {KJ}/{mol} = -ve$$
We also know that activation enrgy always has a positive value and is always at a higher energy level than the reactants or products.
Hence for the above reaction, the correct pathway diagram is $$\left( D \right)$$.
Question 3
1 / -0

If reaction A and B are given with Same temperature and same concentration but rate of $$A$$ is double than $$B$$. Pre exponential factor is same for both the reaction then difference in activation energy $$E_{A} - E_{B}$$ is?

A
$$-RT\ ln2$$

B
$$RT\ ln2$$

C
$$2RT$$

D
$$\dfrac {RT}{2}$$

Solution

$$\dfrac {r_{A}}{r_{B}} = \dfrac {A_{1}e^{\dfrac {-E_{A}}{RT}}}{A_{2}e^{\dfrac {-E_{B}}{RT}}}$$
$$\dfrac {2}{1} = \dfrac {e^{\dfrac {-E_{A}}{RT}}}{e^{\dfrac {-E_{B}}{RT}}}$$
$$ln2 = E_{B} - E_{A} / RT$$
$$E_{B} - E_{A} = RT\ ln2$$
$$E_{A} - E_{B} = -RT\ ln2$$.
Question 4
1 / -0

For what type of the following reactions is the law of mass action, never obeyed?

A
Zero order

B
First order

C
Second order

D
Third order

Solution

In law of mass action molecularity becomes equal to order, which is not possible in case of zeroth order reaction because molecularity can never be zero.
Question 5
1 / -0

$$k$$ for a zero order reaction is $$2\times 10^{-2}\ mol\ L^{-1}\ s^{-1}$$. If the concentration of the reactant after $$25\ s$$ is $$0.5\ M$$, the initial concentration must have been:

A
$$0.5\ M$$

B
$$1.25\ M$$

C
$$12.5\ M$$

D
$$1.0\ M$$

Solution

$$K=2\times 10^{-2}mole/lit-sec$$

At $$t=25\ sec$$. concentration $$=0.5\ M$$

At $$=Ao-Kt$$

$$0.5=Ao-2\times 10^{-2}\times 25\Rightarrow Ao=1\ M$$
Question 6
1 / -0

$$10\%$$ of a reactant decomposes in $$1$$ hour , $$20\%$$ in $$2$$ hours and $$30\%$$ in $$3$$ hours. The order of the reaction is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Given reaction is a zero order reaction because on plotting the percent decomposition of reaction as time, we will obtain a straight line. A straight line is a characteristic of a zero order reaction.
Question 7
1 / -0

In a first order reaction of the type $$A(g)\rightarrow 2B(g)$$, the initial and final pressures are $$p_{1}$$ and $$p$$ respectively. The rate constant can be expressed by

A
$$k=\dfrac{1}{t}\ln \dfrac{p_{1}}{2p_{1}-p}$$

B
$$k=\dfrac{1}{t}\ln =dfrac{p_{1}}{p_{1}-p}$$

C
$$k=\dfrac{1}{t}\ln \dfrac{p_{1}}{p-p_{1}}$$

D
$$k=\dfrac{1}{t}\ln \dfrac{p_{1}}{p}$$

Solution
Question 8
1 / -0

Half-life is independent of the concentration of $$A$$. After $$10$$ minutes, the volume of $$N_2$$ gas is $$10 L$$ and after completion of reaction, it is $$50 L$$.
Hence, rate constant is:

A
$$\dfrac{2.303}{10}\log 5$$ min$$^{-1}$$

B
$$\dfrac{2.303}{10}\log {1.25}$$ min$$^{-1}$$

C
$$\dfrac{2.303}{10} \log 2$$ min$$^{-1}$$

D
$$\dfrac{2.303}{10}\log 4$$ min$$^{-1}$$

Solution

$$t_{50\%}$$ is independent of the concentration of $$A$$. Hence first-order reaction.

$$A$$ $$\rightarrow N_{2}(g)$$

At $$\mathrm{t}=0$$, $$a =0$$

At time $$\mathrm{t},(\mathrm{a}-\mathrm{x}); \ \ \ \ \ \mathrm{x}=$$ $$10 L$$
(after 10 minutes)

At complete reaction, $$a$$ $$=$$ $$50 L$$

$$\therefore \left ( a-x \right )=40L$$

$$\therefore k=\dfrac{2.303}{10}log\dfrac{50}{40}$$

$$k=\dfrac{2.303}{10}log(1.25)\ min^{-1}$$

Option B is correct.
Question 9
1 / -0

The gas phase decomposition of dimethyl ether follows first order kinetics:
$$CH_{3}-O-CH_{3}(g)\rightarrow CH_{4}(g)+H_{2}(g)+CO(g)$$
The reaction is carried out in a constant volume container at $$50^0 C$$ and has a half life of 14.5 minutes. Initially, only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure of the system after 12 minutes? (Assume the ideal gas behaviour.)

A
0.946 atm

B
0.785 atm

C
0.777 atm

D
0.749 atm

Solution

$$t_{\frac{1}{2}}=\dfrac{0.693}{k}\Rightarrow k=\dfrac{0.693}{14.5}=0.0478$$
$$P_t=P_o[e^{kt}]$$
$$P_t = 0.225$$
$$CH_3OCH_3\rightarrow CH_4+H_2+CO$$
$$\begin{matrix}f=0 &\ .40 &0\ \ \ \ &0\ \ \ &0 \\f=t \ &.40-x &x &x &x \end{matrix}$$
$$0.40 - x = 0.225$$
$$x = 0.40 - 0.225 = 0.17 46$$ atm
total pressure$$ = 0.40 + 2x$$
$$= 0.40 + 0.349$$
$$=0.749$$ atm
Question 10
1 / -0

If the initial pressure of $$CH_{3}CHO(g)$$ is 80 mm and the total pressure at the end of 20 min is 120 mm.

$$CH_{3}CHO(g)\rightarrow CH_{4}(g)+CO(g)$$

What is the half-life of the first-order reaction?

A
80 min

B
120 min

C
20 min

D
40 min

Solution

Let $$p^o$$ be the initial pressure. $$p^o=80.$$

At time t, the total pressure is $$\mathrm{p}^{o}+\mathrm{p}=120.$$

$$\mathrm{p}=120-80=40.$$

The expression for the rate constant is $$\mathrm{K}=\cfrac{1}{\mathrm{t}}\ln\cfrac{\mathrm{p}^{o}}{\mathrm{p}^{o}-\mathrm{p}}$$

$$=\cfrac{1}{20}\ln\cfrac{80}{80-40}=\cfrac{1}{20}\ln 2.$$

The expression for the half life period is $$\displaystyle \mathrm{t}_{1/2}=\frac{\ln 2}{\mathrm{K}}.$$

Hence, $$\displaystyle \mathrm{t}_{1/2}=\frac{\ln 2}{\ln 2}\times 20=20\min.$$