Chemical Kinetics Test

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Chemical Kinetics Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

99% of a first order reaction was completed in 32 min. When will 99.9% of the reaction complete ?

A
24 min

B
8 min

C
4 min

D
48 min

Solution

$$k=\frac {2.303}{32} log\frac {100}{1}$$

$$\displaystyle \therefore t=\frac {2.303}{\frac {2.303}{32}log 100} log\frac {100}{0.1}$$

$$\Rightarrow t=48\ min$$.
Question 2
1 / -0

For the first order reaction, $$A(g)\rightarrow 2B(g)+C(g)$$, the initial pressure is $$P_A=90$$ mm Hg. Then pressure after 10 minutes is found to be 180 mm Hg. The half-life period of the reaction is:

A
$$1.15\times 10^{-3} sec^{-1}$$

B
$$600\;sec$$

C
$$3.45\times 10^{-3} sec^{-1}$$

D
$$200\;sec$$

Solution

$$A\rightarrow 2B+C$$
$$P$$ $$0$$ $$0$$
$$P-x$$ $$2x$$ $$x$$
Total pressure after 10 minute is $$180=P-x+2x+x$$
$$180=90+2x$$
$$2x=90, x=45$$

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

$$=\dfrac {2.303}{10} log\dfrac {90}{90-45}$$

$$=\dfrac {2.303}{10\times 60} log2$$

$$=1.1555\times 10^{-3} sec^{-1}$$
$$\therefore t_{\frac {1}{2}}=0.692/k$$
$$ \approx 600 sec$$
Question 3
1 / -0

In a hypothetical reaction, $$A(aq) \rightleftharpoons 2B(aq) + C(aq)$$ (1$$^{st}$$ order decomposition)
'A' is optically active (dextro-rotatory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with $$H_2O_2$$. Hence the progress of reaction can be monitored by measuring rotation of plane of polarised light or by measuring volume of $$H_2O_2$$ consumed in titration.

In an experiment, the optical rotation was found to be $$\theta = 30^o$$ at $$ t = 20$$ min. and $$\theta = 15^o$$ at $$t = 50$$ min. from start of the reaction. If the progress would have been monitored by titration method, volume of $$H_2O_2$$ consumed at $$t=30$$. If the progress would have been monitored by titration method, volume of $$H_2O_2$$ consumed at $$t = 30$$ min. (from start) is 30 ml then volume of $$H_2O_2$$ consumed at $$t = 90$$ min. will be:

A
60 ml

B
45 ml

C
52.5 ml

D
90 ml

Solution

Only A is optically active.
Hence, the concentration of A at $$t = 20\ min \ \propto 30^o$$ and the concentration of A at $$t = 50 \ min \propto 15^o$$

Thus in 30 min, the concentration decreases to half. So 30 min is the half-life period.

Thus, at $$t = 30\ min = t_{1/2}$$, the volume of hydrogen peroxide consumed corresponds to 50% production of B.

$$t = 90$$ min corresponds to three 30 min. The production of B is 87.5%.

Thus volume consumed $$(30\ ml) + \left ( \frac{30}{2}\ ml \right ) + \left ( \frac{30}{4} \right )\ ml = 52.5\ ml$$
Question 4
1 / -0

Directions For Questions

The rate law for the decomposition of gaseous $$N_2O_5$$ is $$N_2O_5\rightarrow 2NO_2+\dfrac {1}{2}O_2$$.
Reaction mechanism has been suggested as follows:

$$N_2O_5 \overset{K_{eq}}\leftrightharpoons NO_2 +NO_3 \ \ \ (fast \ equilibrium)$$

$$NO_2 + NO_3 \xrightarrow {k_1} NO_2 + NO + O_2 \ \ \ (slow)$$

$$NO + NO_3 \xrightarrow {k_2} 2NO_2 \ \ \ (fast)$$

...view full instructions

In 20 minutes of 80% of $$N_2O_5$$ is decomposed. Rate constant is:

A
0.08

B
0.05

C
0.12

D
0.2

Solution

$$k=\frac {2.303}{t} log \frac {A_0}{A_t}=\frac {2.303}{20} log \frac {100}{20}$$
On solving $$k=0.08$$
Question 5
1 / -0

The inactivation of a viral preparation in a chemical bath is found to be a first-order reaction. The rate constant for the viral inactivation per minute, if in the beginning $$1.5\%$$ of the virus is inactivated, is:

A
$$1.25\times 10^{-4}\ sec^{-1}$$

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

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

D
$$2.5\times 10^{-4}\ min^{-1}$$
Question 6
1 / -0

The high temperature $$(\approx 1200 K)$$ decomposition of $$CH_3 COOH (g)$$ occurs as follows, as per simultaneous 1$$^{st}$$ order reactions.
$$CH_3 COOH \xrightarrow{K_1} CH_4 + CO_2$$
$$CH_3 COOH \xrightarrow{K_2} CH_2 CO + H_2O$$
What would be the % of CH$$_4$$ by mole in the product mixture (excluding $$CH_3COOH$$)?

A
$$\displaystyle \frac{50k_1}{(k_1 + k_2)}$$

B
$$\displaystyle \frac{100k_1}{(k_1 + k_2)}$$

C
$$\displaystyle \frac{200k_1}{(k_1 + k_2)}$$

D
It depends on time

Solution

Let $$V$$ be the total volume of the mixture.
The equilibrium constant expression for the first reaction is $$\displaystyle k_1 = \frac {n_{CH_4}n_{CO_2}}{n_{CH_3COOH} \times V} $$......(1)
The equilibrium constant expression for the second reaction is $$\displaystyle k_2 = \frac {n_{CH_2CO}n_{H_2O}}{n_{CH_3COOH} \times V} $$......(2)
The ratio of the equilibrium constants for two reactions is obtained by dividing equation (1) with equation (2). It is
$$\displaystyle \frac{^n CH_4 + ^nCO_2}{^nCH_2 CO + ^n H_2O} = \frac{k_1}{k_2}$$.

Hence, the value of the term $$ \dfrac{k_1}{k_1 + k_2}$$ is $$\displaystyle \frac{^nCH_4 + ^nCO_2}{^nCH_4 + ^nCO_2 + ^nCO_2CO + ^nH_2O} = \frac{k_1}{k_1 + k_2}$$

or $$ \displaystyle \frac{^{2N}CH_4}{n_{total}} = \frac{k_1}{k_1 + k_2}$$

$$\displaystyle \frac{^nCH_4}{n_{total}} = \frac{k_1}{2(k_1 + k_2)}$$

$$\displaystyle \frac{^nCH_4}{n_{total}} \times 100 = \frac{50 k_1}{(k_1 + k_2)}$$.

Hence, then the percentage of methane by mole in the product mixture (excluding acetic acid) is $$\displaystyle \frac{50k_1}{(k_1 + k_2)}$$.
Question 7
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For the first order homogeneous gaseous reaction $$A\rightarrow 2B+C$$, the initial pressure was $$P_i$$ while total pressure at time 't' was $$P_t$$. Write expression for the rate constant k in terms of $$P_i, P_t$$ & $$t$$.

A
$$k=\cfrac {2.303}{t}log \left (\cfrac {2P_i}{3P_i-P_t}\right )$$

B
$$k=\cfrac {2.303}{t}log \left (\cfrac {2P_i}{ 2P_t-P_i}\right )$$

C
$$k=\cfrac {2.303}{t}log \left (\cfrac {P_i}{P_i-P_t}\right )$$

D
None of these

Solution

The first-order reaction is $$A\rightarrow 2B+C$$

A
B
C
At time 0: Pi
0
0
At time t:
Pi−x 2x
x

Hence, the expression for the rate constant will be:

$$K=\dfrac {2.303}{t}log \left(\cfrac {P_i}{P_i-x}\right)$$

But $$P_i-x+2x+x=P_t$$

Hence, $$x=\cfrac {1}{2}(P_t-P_i)$$ or $$P_i-x=\cfrac {1} {2}(3P_i-P_t)$$

Thus the expression for the rate constant becomes

$$K=\dfrac {2.303}{t} log \left(\cfrac {2P_i}{(3P_i-P_t)}\right)$$

Hence, the correct option is $$\text{A}$$
Question 8
1 / -0

Compounds A and B react with a common reagent with first order kinetics in both cases. If 99% of A must react before 1% of B has reacted, what is the minimum ratio for their respective rate constants?

A
$$916$$

B
$$229$$

C
$$500$$

D
$$458$$

Solution

The fraction of A and B remaining at time t when 99% of A has reacted, are (A $$=$$ 0.01, B $$=$$ 0.00)

$$\displaystyle k_A = \frac{1}{t} \ln\frac{100}{1}$$

$$\displaystyle k_B = \frac{1}{t} \ln\frac{100}{99}$$

$$\displaystyle \therefore \frac{k_A}{k_B} = \frac{\ln 100}{\ln \left[\cfrac{100}{99}\right]} =458$$

Hence, the correct option is $$\text{D}$$
Question 9
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For the following data for the zero order reaction $$A\rightarrow$$ products. Calculate the value of k.
Time [A]
0.0 0.10 M
1.0 0.09 M
2.0 0.08 M

A
$$k=0.01 \, M min^{-1} $$

B
$$k=0.03 \, M min^{-1} $$

C
$$k=0.06 \, M min^{-1} $$

D
$$k=0.08 \, M min^{-1} $$

Solution

For zero order reaction:
$$rate = k$$

$$k = (0.10-0.09)/1 = 0.01/1 = 0.01 M min^{-1}$$

$$Rate = \dfrac{0.09-0.08}{2-1}= 0.01 M min^{-1}$$
Question 10
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Directions For Questions

The rate law expresses the relationship between the rate of a reaction and the rate constant and the concentrations of the reactants raised to some powers. For the general reaction:
$$aA+bB \rightarrow cC + dD$$
Rate law takes the form
$$r = k [A]^x [B]^y$$
where $$x$$ and $$y$$ are numbers that must be determined experimentally, $$k$$ is the rate constant and $$[A]$$ and $$[B]$$ are a concentration of $$A$$ and $$B$$ respectively.

...view full instructions

Gaseous reaction $$A \rightarrow B + C$$ follows first-order kinetics concentration of $$A$$ changes from $$1\ M$$ to $$0.25\ M$$ in 138.9 min. Find the rate of reaction when conc. of $$A$$ is $$0.1\ M$$.

A
$$10^{-2} M \ min^{-1}$$

B
$$10^{-3} M\ min^{-1}$$

C
$$10^{-4} M\ min^{-1}$$

D
$$10^{-5} M\ min^{-1}$$

Solution

$$\displaystyle r = k [A] $$
$$= \frac{0.693}{69.3} \times 0.1 $$
$$= 10^{-3} Mmin^{-1}, t_{1/2} $$
$$= 69.3 min.$$

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	A	B	C
At time 0:	Pi	0	0
At time t:	Pi−x	2x	x

Chemical Kinetics Test - 58

Result

Chemical Kinetics Test - 58

Score

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Rank

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SHARING IS CARING

Question 1

24 min

8 min

4 min

48 min

Solution

Question 2

$$1.15\times 10^{-3} sec^{-1}$$

$$600\;sec$$

$$3.45\times 10^{-3} sec^{-1}$$

$$200\;sec$$

Solution

Question 3

60 ml

45 ml

52.5 ml

90 ml

Solution

Question 4

0.08

0.05

0.12

0.2

Solution

Question 5

$$1.25\times 10^{-4}\ sec^{-1}$$

$$2.5\times 10^{-4}\ sec^{-1}$$

$$5\times 10^{-4}\ sec^{-1}$$

$$2.5\times 10^{-4}\ min^{-1}$$

Question 6

$$\displaystyle \frac{50k_1}{(k_1 + k_2)}$$

$$\displaystyle \frac{100k_1}{(k_1 + k_2)}$$

$$\displaystyle \frac{200k_1}{(k_1 + k_2)}$$

It depends on time

Solution

Question 7

$$k=\cfrac {2.303}{t}log \left (\cfrac {2P_i}{3P_i-P_t}\right )$$

$$k=\cfrac {2.303}{t}log \left (\cfrac {2P_i}{ 2P_t-P_i}\right )$$

$$k=\cfrac {2.303}{t}log \left (\cfrac {P_i}{P_i-P_t}\right )$$

None of these

Solution

Question 8

$$916$$

$$229$$

$$500$$

$$458$$

Solution

Question 9

$$k=0.01 \, M min^{-1} $$

$$k=0.03 \, M min^{-1} $$

$$k=0.06 \, M min^{-1} $$

$$k=0.08 \, M min^{-1} $$

Solution

Question 10

$$10^{-2} M \ min^{-1}$$

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

$$10^{-4} M\ min^{-1}$$

$$10^{-5} M\ min^{-1}$$

Solution

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