Chemical Kinetics Test

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Chemical Kinetics Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

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.

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The initial rate of zero-order reaction of the gaseous equation $$A$$ (g) $$\rightarrow$$ $$2B$$ (g) is 10$$^{-2}$$ M min$$^{-1}$$ if the initial concentration of $$A$$ is 0.1 M. What would be a concentration of $$B$$ after $$60$$ seconds?

A
$$0.09 M$$

B
$$0.01 M$$

C
$$0.02 M$$

D
$$0.03 M$$

Solution

For a zero order reaction, the rate law expression is $$R = k[A]^0=k=10^{-2}$$
$$t=60 s = 1 min$$
$$[A_{0}] - [A] = Kt$$
$$0.1 - [A] = 10^{-2} \times 1$$
$$[A]=0.09 M$$
$$[B]=2[A]$$
$$=2(0.1-0.09)$$
$$= 0.02 M$$
Question 2
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A certain substance A is mixed with an equimolar quantity of substance B. At the end of an hour, A is 75% reacted. Calculate the time when A is 10% unreacted. (Given: order of reaction is zero)

A
1.2 hr

B
1.8 hr

C
2.0 hr

D
2.4 hr

Solution

According to zero order reaction:
$$x =kt$$ ( x = amount of reactant reacted)

$$ 75 = k\times1$$ and
$$90 = k\times t$$
So $$t= 90/75 =1.2$$ hr
Question 3
1 / -0

Time t $$\infty $$
Rotation of Glucose & Fructose $$ r_{t} $$ $$r _{\infty }$$
$$S\rightarrow$$ $$G+F $$
What id the value of $$k$$ for the above reaction under given circumstances?

A
$$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{\infty }-r_{t})}$$

B
$$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{\infty }-r_{t})}$$

C
$$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{t}-r_{\infty})}$$

D
$$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{t}-r_{\infty})}$$

Solution

The rate law expression for the first order reaction is $$k=\frac {2.303} {t}log \frac {a} {a-x}$$.

But $$a \propto r _{\infty } $$ and $$ x \propto r_{t}$$

Substitute values in the above equation.

$$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{\infty }-r_{t})}$$
Question 4
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A metal slowly forms an oxide film which completely protects the metal when the film thickness is 3.956 thousandths of an inch. If the film thickness is 1.281 thou. in 6 weeks, then the time it will take before it is 2.481 thou. is :(the rate of film formation follows first-order kinetics)

A
12 weeks

B
14 weeks

C
15 weeks

D
17 weeks

Solution

When the film thickness is 3.956 thousandths of an inch., $$a=\frac {3.956} {1000}$$.
For $$t=6 weeks, a-x=\frac {(3.956-1.281)} {1000}$$
$$k=\frac {2.303} {t} log \frac {a}{a-x}=\frac {2.303} {6} log \frac {3.956}{3.956-1.281}$$......(1)
For $$t=? weeks$$, $$a-x=\frac {(3.956-2.481)} {1000}$$
$$k=\frac {2.303} {t} log \frac {a}{a-x}=\frac {2.303} {t} log \frac {3.956}{3.956-2.481}$$......(2)
From equations (1) and (2),
$$\frac {2.303} {6} log \frac {3.956}{3.956-1.281}=\frac {2.303} {t} log \frac {3.956}{3.956-2.481}$$
Hence, $$t=12 weeks$$.
Question 5
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At $$100^{\circ}$$C the gaseous reaction $$A\rightarrow $$2B+C was observed to be of first order. On starting with pure A it is found that at the end of 10 minutes the total pressure of system is 176mm.Hg and after a long time 270mm Hg.
Initial pressure of A is:

A
90mm

B
80mm

C
70mm

D
60mm

Solution

After long time the dissociation of A is complete. The dissociation of one mole of A gives 3 moles of products. At this time, total pressure is 270mm Hg. Hence, the initial pressure of $$A=\frac{270}{3}=90mm$$.
Question 6
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The rate constant for disappearance of reactant as mentioned in is $$\displaystyle 2 \times 10^{-2} mol L^{-1} sec^{-1}$$, if the concentration of the reactant after 25 sec is 0.25M, the initial concentration is:

A
0.75 M

B
0.89 M

C
1.03 M

D
1.46 M

Solution

From rate constant it can be seen that it is zero order reaction and $$a-x = kt;$$ $$a = x+kt$$ $$= 0.25+2 \times 10^{-2} \times 25 = 0.75$$ M.
Question 7
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For a reversible first - order reaction $$A\rightleftharpoons B K_{1}=10^{-2}s^{-1}$$ and $$[B]_{eq}=4$$. If $$[A]_{0}=0.01$$ mole $$L^{-1} $$ and $$ [B]_{0}=0$$, what will be the concentration of B after 30 s?

A
0.0025 m

B
0.0013 m

C
0.0026 m

D
0.0030 m

Solution

As we know,
$$k_1+k_2 = (1/t)log(x_e - x_0)/(x_e - x)$$ and
$$k_e = k_1/k_2$$
And from given data, conc. of B after 30 sec.
$$k_1+k_2 = 1/30)log(4-0/4-x)$$
$$x= 0.0025$$m
Question 8
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In this case, we have (Consider it as a first-order reaction)
$$A\rightarrow B+C$$
Time $$t$$ $$\infty $$
Total pressure $$ p_{2}$$ $$ \ \ \:\:\:\:p_{3}$$
Then $$k$$ is:

A
$$\displaystyle k=\frac{1}{t}ln\frac{p_{3}}{2(p_{3}-p_{2})}$$

B
$$\displaystyle k=\frac{1}{t}ln\frac{p_{2}}{2(p_{3}-p_{2})}$$

C
$$\displaystyle k=\frac{1}{t}ln\frac{p_{3}}{2(p_{1}-p_{2})}$$

D
$$\displaystyle k=\frac{1}{t}ln\frac{p_{1}}{2(p_{3}-p_{2})}$$

Solution

When $$t=\infty$$, entire $$A$$ has reacted. 1 mole of $$A$$ gives 1 mole of $$B$$ and 1 mole of $$C$$ $$a=\dfrac {p_3} {2}$$.
$$t=t$$, the total pressure is $$a-x+x+x=a+x=p_2$$.
Hence, $$x=p_2-a$$.
$$a-x=a-(p_2-a)= 2a-p_2=p_3-p_2$$
The integrated rate law for the first order reaction is $$k=\dfrac {1} {t} ln{\dfrac {[A]_0} {[A]_t}}=\dfrac {1} {t} ln {\dfrac {P_3} {2}}/( {P_3-P_2})=\displaystyle k=\dfrac{1}{t}ln\dfrac{p_{3}}{2(p_{3}-p_{2})}$$
Question 9
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$$\displaystyle SO _{2}Cl _{2}(g)\rightarrow SO_{2}(g)$$

The given reaction is a first order gas reaction with $$k=2.2 \times \displaystyle 10^{-5}sec^{-1}$$ at $$ 320^{\circ}$$C. What % of $$SO_{2}Cl_{2}$$ is decomposed on heating this gas for 90 min?

A
11.2%

B
12.3%

C
13.4%

D
14.5%

Solution

for a first order reaction:

$$k = 2.303\ log\left(\cfrac{\frac{a}{a-x}}{t}\right);$$

$$ k=2.2\times 10^{-5} = 2.303log\left(\cfrac{\frac{a}{a-x}}{90 \times 60}\right)$$

so $$\cfrac{(a-x)\times100 }a = 88.8$$ %

so % decomposed $$= 100-88.8= 11.2$$ %

Hence, the correct option is $$\text{A}$$
Question 10
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A vessel contains dimethy 1 ether at a pressure of 0.4 atm. Dimethyl ether decomposes as $$ CH_{3}OCH_{3}(g)\rightarrow CH_{4}(g)+CO(g)+H_{2}(g).$$ The rate constant of decomposition is $$4.78\times10^{-3}min.$$ The ratio of initial rate of diffusion after 4.5 hours of initiation of decomposition is:

A
0.26 : 1

B
1 : 0.26

C
0.34 : 2

D
2 : 0.34

Solution

The pressure after 4.5 hours is $$P=P_oe^{-kt}=0.4e^{-4.78 \times 10^{-3} \times 4.5 \times 60}=0.11 atm$$.
The unit of the rate constant indicates first order reaction.
The expression for the rate of the reaction is $$r=kp$$.
The ratio of initial rate of diffusion after 4.5 hours of initiation of decomposition is $$\dfrac {r_t} {r_o} =\dfrac {P_t} {P_o}=\dfrac {0.11} {0.4}=\dfrac {0.26} {1}$$.

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Time	t	$$\infty $$
Rotation of Glucose & Fructose	$$ r_{t} $$	$$r _{\infty }$$

Chemical Kinetics Test - 59

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Chemical Kinetics Test - 59

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SHARING IS CARING

Question 1

$$0.09 M$$

$$0.01 M$$

$$0.02 M$$

$$0.03 M$$

Solution

Question 2

1.2 hr

1.8 hr

2.0 hr

2.4 hr

Solution

Question 3

$$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{\infty }-r_{t})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{\infty }-r_{t})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{t}-r_{\infty})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{t}-r_{\infty})}$$

Solution

Question 4

12 weeks

14 weeks

15 weeks

17 weeks

Solution

Question 5

90mm

80mm

70mm

60mm

Solution

Question 6

0.75 M

0.89 M

1.03 M

1.46 M

Solution

Question 7

0.0025 m

0.0013 m

0.0026 m

0.0030 m

Solution

Question 8

$$\displaystyle k=\frac{1}{t}ln\frac{p_{3}}{2(p_{3}-p_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{p_{2}}{2(p_{3}-p_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{p_{3}}{2(p_{1}-p_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{p_{1}}{2(p_{3}-p_{2})}$$

Solution

Question 9

11.2%

12.3%

13.4%

14.5%

Solution

Question 10

0.26 : 1

1 : 0.26

0.34 : 2

2 : 0.34

Solution

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