Chemistry Case study (4 Marks)

1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

Rate expression $\frac{\text{dx}}{\text{dt}}=\text{k}[\text{A}]^\text{m}[\text{B}]^\text{n}$

shows that the total order of reactions is m + n + 0 = m + n, as the rate of reaction is independent of concentration of C, i.e., the order with respect to C is zero. This is the reason that C does not figure in the rate expression.

2. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

Instantaneous rate of a reaction is equal to small change in concentration (dx) during a small interval of time (dt) at that particular instant of time divided by the time interval.

3. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

For the given reaction, rate of reaction (r) = k[RCl] (where k is rate constant). Therefore if the concentration of [RCl] is reduced to half, then new rate $(\text{r}')=\frac{\text{k}}{2}[\text{RCl}].$

4. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

5. (d) Assertion is wrong statement but reason is correct statement.

Explanation:

2A + B → A₂B

The rate of disappearance of reactant will be

$-\frac{1}{2}\frac{\text{d[A]}}{\text{dt}}=\frac{\text{d[B]}}{\text{dt}}=\frac{\text{d}[\text{A}_2\text{B}]}{\text{dt}}$

Therefore, the reactant B will disappear at half the rate as A will decrease.

1. (c) 0.01s^-1

Explanation:

$\text{t}_\frac{1}{2}=69.3\text{s}$

For first order reaction,

$\text{k}=\frac{0.693}{\text{t}_\frac{1}{2}}=\frac{0.693}{69.3}=0.01\text{s}^{-1}$

2. (a) 0.04 atm

Explanation:

$\text{k}=\frac{2.303}{\text{t}}\log\frac{\text{a}}{\text{a}-\text{x}}=\frac{2.303}{230}\log\frac{0.4}{0.4-\text{x}}$

$0.01=\frac{2.303}{230}\log\frac{0.4}{0.4-\text{x}}=0.01\log\frac{0.4}{0.4-\text{x}}$

$\frac{0.01}{0.01}=\log\frac{0.4}{0.4-\text{x}}\Rightarrow1=\log\frac{0.4}{0.4-\text{x}}$

Antilog of $1=\frac{0.4}{0.4-\text{x}}\Rightarrow10=\frac{0.4}{0.4-\text{x}}\Rightarrow\text{x}=0.36$

$\therefore\text{a}-\text{x}=0.4-0.36=0.04\text{ atm}$

3. (b) 1.12 atm

Explanation:

For the given reaction,

$\text{A}_{(\text{g})}\xrightarrow{\ \ \triangle\ \ \ }\text{P}_{(\text{g})}+\text{Q}_{(\text{g})}+\text{R}_{(\text{g})}$

Total pressure = (0.4 - 0.36) + (3 × 0.36) = 1.12 atm

4. (b) Linear with intercept = In[A]₀

Explanation:

Expression that relates concentration of reactant and ti me for first order reaction is

In[A] = -kt + In[A]₀

So, the plot of In [A] vs twill be linear with slope = -k and intercept = In[A]₀

5. (c) $2\text{N}\text{H}_{3(\text{g})}\xrightarrow[\triangle]{\text{pt}}\text{N}_{2(\text{g})}+3\text{H}_{2(\text{g})}$

Explanation:

Decomposition of ammonia on a hot platinum surface at high pressure is a zero order reaction.

1. (c) Low probability of simultaneous collision of all the reacting species.

Explanation:

The reactions of higher order are very rare because of the less chances of the molecules to come together simultaneously and collide.

2. (c) 3

Explanation:

The total number of reactant molecules participating in a chemical reaction is known as its molecularity, hence the molecularity = 6 + 3 + 1 = 10.

3. (c) Molecularity and order always have same values for a reaction.

Explanation:

Molecularity may or may not be equal to the order of a reaction.

4. (b) First

Explanation:

From the expression

$-\frac{\text{d}[\text{A}]}{\text{dt}}=\text{k}[\text{A}][\text{B}]^2$

when B is present in large excess, rate will be independent upon the change in cone. of B, therefore order of reaction will be one.

5. (d) $\frac{13}{12}$

Explanation:

Order of reaction $=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{6+4+3}{12}=\frac{13}{12}$

1. (a) 0.693 × 3M min^-1

Explanation:

Unit of rate constant for a reaction of n^th order= (conc.)^{1 - n} time^-1

For a first order reaction, n = 1

Unit of rate constant = (mol L^-1)^1-1 s^-1 = s^-1

2. (d) 0.0693 × 3M min^-1

Explanation:

$12\text{M}\xrightarrow{\ \ \ \text{t}_\frac{1}{2}\ \ \ \ }6\text{M}\xrightarrow{\ \ \ \text{t}\frac{1}{2}\ \ \ }3\text{M}\\_\text{initial conc}$

$\text{t}_\frac{1}{2}=10\text{min}$

$\text{k}=\frac{0.693}{10}=0.0693\text{min}^{-1}$

As $\text{t}_\frac{1}{2}$ is 10 min, after 20 minutes the concentration will be 3M.

Hence, Rate = 0.0693 × 3M min^-1

3. (b) 76.4 minutes.

Explanation:

$\text{t}_\frac{1}{2}=23\text{ minutes}$

$\text{t}_\frac{1}{2}=\frac{0.693}{\text{k}}\Rightarrow\text{k}=\frac{0.693}{\text{t}_\frac{1}{2}}$

$\Rightarrow\frac{0.693}{23}\text{min}^{-1}$

For 90% completion,

$\text{t}=\frac{2.303}{\text{k}}\log\Big(\frac{\text{a}}{\text{a}-\text{x}}\Big)$

$\text{t}=\frac{2.303\times23}{0.693}\log\Big(\frac{100}{100-90}\Big)$

$\text{t} = 76.4\text{ minutes}$

4. (a) 3.47 × 10^-4 M/ min

Explanation:

For the first order reaction,

$\text{k}=\frac{2.303}{\text{t}}\log\frac{\text{a}}{\text{a}-\text{x}}$

$\text{a}=0.1\text{M},\text{a}-\text{x}=0.025\text{M},\text{t}=40\text{min}$

$\text{k}=\frac{2.303}{40}\log\frac{0.1}{0.025}=\frac{2.303}{40}\log4$

$=0.0347\text{min}^{-1}$

[A] → product

Thus, rate = k[A]

rate = 0.0347 × 0.01M min^-1 = 3.47 × 10^-4M min^-1

5. (a) 6.25%

Explanation:

$\text{t}_\frac{1}{2}=\frac{0.693}{\text{k}}\Rightarrow\frac{0.693}{\text{t}_\frac{1}{2}}=\text{k}\Rightarrow\frac{0.693}{60}=\text{k}$

$\text{k} = 0.01155 \text{ min}^{-1}$

$\text{k}=\frac{2.303}{\text{t}}\log\Big(\frac{\text{a}}{\text{a}-\text{x}}\Big)$

Let the initial amount (a) be 100.

$0.01155\text{min}^{-1}=\frac{2.303}{240\text{min}}\log\Big(\frac{100}{\text{a}-\text{x}}\Big)$

$\frac{0.01155^{-1}\times240\text{min}}{2.303}=\log\Big(\frac{100}{\text{a}-\text{x}}\Big)$

$1.204=\log100-\log(\text{a}-\text{x})$

$1.204=2-\log(\text{a}-\text{x})$

$\log(\text{a}-\text{x})=2-1.204=0.796$

$(\text{a}-\text{x})=6.25\%$

1. (b) 2

Explanation:

According to the figure, in the given time of 4 hours ( 1 to 5) concentration of A falls from 0.5 to 0.3M, while in the same time concentration of B increases from 0.2 to 0.6 M.

Decrease in concentration of A in 4 hours

= 0.5 - 0.3 = 0.2M

Increase in concentration of B in 4 hours

= 0.6 - 0.2 = 0.4M

Thus, increase in concentration of B in a given time is twice the decrease in concentration of A. Thus, n = 2.

2. (b) 1.2M

Explanation:

$\text{K}=\frac{[\text{B}]^2}{[\text{A}]}=\frac{(0.6)^2}{0.3}=1.2\text{M}$

3. (a) 0.1 mol L^-1hr^-1

Explanation:

From t = 0 to t = 1hr,

For A, dx = 0.6 - 0.5 = 0.1 mol L^-1

$\therefore$ Initial rate of conversion of $\text{A}=\frac{\text{dx}}{\text{dt}}$

$=\frac{0.1\text{ mol}\text{ L}^{-1}}{1\text{hr}}=0.1\text{ mol}\text{ L}^{-1}\text{hr}^{-1}$

4. (b) 10^-4

Explanation:

$\text{A}\rightleftharpoons2\text{B}$

$-\frac{\text{d}[\text{A}]}{\text{dt}}=+\frac{1}{2}\frac{\text{d}[\text{B}]}{\text{dt}}$

$=\frac{1}{2}\times2\times10^{-4}=10^{-4}$

5. (d) Molecularity.

Explanation:

The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction is called molecularity and it has no influence on the rate of reaction.

1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

2. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

The reaction proceeds with a constant rate which is independent of concentration of hydrogen and chlorine. That is why, this reaction is a zero order reaction.

3. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

For a zero order reaction, $\text{t}_\frac{1}{2}=\frac{[\text{a}]}{2\text{k}}$.

4. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

5. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

For a zero order reaction, units of k is mol L^-1 s^-1, i.e., it depends upon units of concentration.

1. (c) 230.3s

Explanation:

For a first order reaction:

$\text{t}_\frac{1}{2}=\frac{0.693}{\text{k}},\text{k}=3.01\times10^{-3}\text{s}^{-1}$

$\therefore\text{t}_\frac{1}{2}=\frac{0.693}{3.01\times10^{-3}}=230.3\text{s}$

2. (a) 990s

Explanation:

Half life $\Big(\text{t}_\frac{1}{2}\Big)$ of a first order reaction is given as:

$\text{t}_\frac{1}{2}=\frac{0.693}{7.0\times10^{-4}}=990\text{s}$

3. (b) It is independent of temperature.

Explanation:

For a first order reaction $\text{t}_\frac{1}{2}=\frac{0.693}{\text{K}}$ therefore $\text{t}_\frac{1}{2}$ depends upon k and hence depends on temperature because rate constant k is a function of temperature.

4. (c) 24.086 min

Explanation:

Let the concentrations of the reactant after 10 min and 20 min be C₁ and C₂ respectively.

$\therefore$ Rate after 10 min = k.C₁

= 0.04 × 60 mol L^-1 min^-1

and rate after 20 min = k.C₂

= 0.03 × 60 mol L^-1 min^-1

$\therefore\frac{\text{C}_1}{\text{C}_2}=\frac{4}{3}$

Let the reaction starts after 10 minutes.

$\text{k}=\frac{2.303}{10}\log\frac{\text{C}_1}{\text{C}_2}=\frac{2.303}{10}\log\frac{4}{3}=0.02878$

$\therefore\text{t}_\frac{1}{2}=\frac{0.6932}{\text{k}}=\frac{0.6932}{0.02878}=24.086\text{min}$

Explanation:

For a first order reactions, $\text{t}_\frac{1}{2}=\text{k[A]}_0^0=\text{k}.$

Thus $\text{t}_\frac{1}{2}$ is independent of initial concentration.

Hence plot of $\text{t}_\frac{1}{2}$ vs [A]₀ will be a horizontal line.

1. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

For first order reaction, Rate, = k[A₁] According to question,

[A₂] = [2A₁]

$\therefore$ Rate2 = k[2A₁]

⇒ Rate₂ = 2 Rate₁

For a given reaction, rate constant is constant and independent of the concentration of reactant.

2. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

For first order reaction $\text{k}=\frac{2.303}{\text{t}}\log\frac{\text{a}}{\text{a}-\text{x}}$

$\text{k}=\frac{2.303}{\text{t}_\frac{1}{2}}\log\frac{\text{a}}{\text{a}-\frac{\text{a}}{2}}=\frac{2.303}{\text{t}_\frac{1}{2}}\log\frac{\text{a}}{\frac{\text{a}}{2}}=\frac{2.303}{\text{t}_\frac{1}{2}}\log2$

Therefore half-life period $\text{t}_\frac{1}{2}=\frac{2.303}{\text{k}}\log2.$

Thus $\text{t}_\frac{1}{2}$ is independent of initial concentration of reactant for first order reaction.

3. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

$\text{CH}_3\text{COOC}_2\text{H}_5+\text{H}_2\text{O}\xrightarrow{\ \ \text{H}^+\ \ }\text{CH}_3\text{COOH}+\text{C}_2\text{H}_5\text{OH}$

Rate $\propto[\text{CH}_3\text{COOC}_2\text{H}_5]$

$\text{CH}_3\text{COOC}_2\text{H}_5+\text{NaOH}\rightarrow\text{CH}_3\text{COONa}+\text{C}_2\text{H}_5\text{OH}$

Rate $\propto[\text{CH}_3\text{COOC}_2\text{H}_5][\text{NaOH}]$

4. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

For a first order reaction, $[\text{A}]=[\text{A}]_0\text{e}^{-\text{kt}}$

or $\log[\text{A}]=-\frac{\text{kt}}{2.303}+\log[\text{A}]_0$

5. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

For a first order reaction, $\text{t}_\frac{1}{2}$ is inversely proportional to k, it does not depend on the initial concentration of the reactant.

1. (c) 3

Explanation:

$2\text{NO}_\text{(g)}+\text{Cl}_{2\text{(g)}}\rightarrow2\text{NOCl}_\text{(g)}$

Molecularity = 3

2. (b) $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]$

Explanation:

Let rate of this reaction, $\text{r}=\text{k}[\text{NO]}^\text{m}[\text{Cl}_2]^\text{n}$

then $\frac{\text{r}_1}{\text{r}_2}=\frac{0.60}{1.20}=\frac{\text{k}(0.15)^\text{m}(0.15)^\text{n}}{\text{k}(0.15)^\text{m}(0.30)^\text{n}}$

or, $\frac{1}{2}=\Big(\frac{1}{2}\Big)^\text{n}\Rightarrow\text{n}=1$

Again from $\frac{\text{r}_2}{\text{r}_3}=\frac{1.20}{2.40}=\frac{\text{k}(0.15)^\text{m}(0.30)^\text{n}}{\text{k}(0.30)^\text{m}(0.15)^\text{n}}$

or $\frac{1}{2}=\Big(\frac{1}{2}\Big)^\text{m}\cdot\frac{2}{1}$ or $\frac{1}{4}=\Big(\frac{1}{2}\Big)^\text{m}\Rightarrow\text{m}=2$

Hence, expression for rate law is

$\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]^1$

3. (d) 3

Explanation:

As the order w.r.t. NO is 2 and order w.r.t. Cl₂ is 1, hence the overall order is 3.

4. (c) 177.77 M^-2 min^-1

Explanation:

Substituting the values of experiment 1 in rate law expression

0.60M min^-1 = k(0.15M)² (0.15M)¹

or $\text{k}=\frac{0.60\text{M min}^{-1}}{0.0225\times0.15\text{M}^3}=177.77\text{M}^{-2}\text{min}^{-1}$

5. (d) 2.77M min^-1

Explanation:

r = 177.7M^-2 min^-1 × (0.25M)² (0.25M)

= 2.77M min^-1

1. (c) 3

Explanation:

Unit of k for n^th order = (mot L^-1)^1-n sec^-1

Here, k = 3 × 10^-3 mol^-2 L² sec^-1

Unit of k = mol^-2 L² sec^-1 ⇒ (mol L^-1)^-2 sec^-1

Comparing (i) and (ii) we get, 1 - n = -2 ⇒ n = 3

2. (c) twice the rate of consumption of A.

Explanation:

Rate $=\frac{\text{d[A]}}{\text{dt}}=-\frac{1}{3}\frac{\text{d[B]}}{\text{dt}}=\frac{1}{2}\frac{\text{d[C]}}{\text{dt}}$

3. (c) 18 times

Explanation:

Given, R₁ = k[A]² [B]

According to question, R₂ = k[3A]² [2B]

= k × 9 [A]² × 2[B] = 18 × k[A]² [B] = 18 R₁

4. (c) 1

Explanation:

Rate (r) = k[H⁺]ⁿ

When pH = 3 ; [H⁺] = 10^-3

and when pH = 1 ; [H⁺] = 10^-1

$\therefore\frac{\text{r}_1}{\text{r}_2}=\frac{\text{k}(10^{-3})^\text{n}}{\text{k}(10^{-1})^\text{n}}\Rightarrow\frac{1}{100}=\Big(\frac{10^{-3}}{10^{-1}}\Big)^\text{n}$ $(\because\text{r}_2=100\text{r}_1)$

$\Rightarrow(10^{-2})^1=(10^{-2})^\text{n}\Rightarrow\text{n}=1$

5. (b) 2

Explanation:

Let the order of reaction w.r.t. A is x and w.r.t. B is y.

$\text{r}_1=\text{k}[\text{A}]^\text{x}[\text{B}]^\text{y}$

$\text{r}_2=\text{k}[\text{2A}]^\text{x}[\text{B}]^\text{y}$

$\text{r}_3=\text{k}[\text{A}]^\text{x}[\text{2B}]^\text{y}$

$\frac{\text{r}_1}{\text{r}_2}=\frac{\text{k}[\text{A}]^\text{x}[\text{B}]^\text{y}}{\text{k}[2\text{A}]^\text{x}[\text{B}]^\text{y}}$

$\Rightarrow\frac{1}{4}=\Big(\frac{1}{2}\Big)^\text{x}\Rightarrow\Big(\frac{1}{2}\Big)^2=\Big(\frac{1}{2}\Big)^\text{x}\Rightarrow\text{x}=2$

Similarly, $\frac{\text{r}_1}{\text{r}_3}=\frac{\text{k}[\text{A}]^\text{x}[\text{B}]^\text{y}}{\text{k}[\text{A}]^\text{x}[2\text{B}]^\text{y}}$

$\Rightarrow1=\Big(\frac{1}{2}\Big)^\text{y}\Rightarrow\Big(\frac{1}{2}\Big)^0=\Big(\frac{1}{2}\Big)^\text{y}\Rightarrow\text{y}=0$

Hence the rate law equation is

Rate = k[A]² [B]⁰ ⇒ Order of reaction = 2