Question 1

1 For the reaction $\mathrm{R} \rightarrow \mathrm{P}$, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Accepted Answer

Given: Initial concentration, [R]₁ = 0.03 M
Final concentration, [R]₂ = 0.02 M
Time interval, t = 25 minutes

Average rate of reaction = change in concentration / time = $\frac{[R]_1 - [R]_2}{t} = \frac{0.03 - 0.02}{25} = \frac{0.01}{25} = 4 \times 10^{-4} \mathrm{M/min} $

To convert time to seconds: 25 minutes = 25 × 60 = 1500 seconds

Average rate in seconds = $\frac{0.01}{1500} = 6.67 \times 10^{-6} \mathrm{M/s} $

Hence, average rate = 4 × 10⁻⁴ M/min or 6.67 × 10⁻⁶ M/s. — The average rate is calculated by dividing the change in concentration by the time taken. First, calculate rate in minutes, then convert time to seconds to find rate in seconds.

Question 2

2 In a reaction, $2\,\mathrm{A} \rightarrow$ Products, the concentration of A decreases from 0.5 mol L$^{-1}$ to 0.4 mol L$^{-1}$ in 10 minutes. Calculate the rate during this interval?

Accepted Answer

Given: Initial concentration, [A]₁ = 0.5 mol L⁻¹
Final concentration, [A]₂ = 0.4 mol L⁻¹
Time interval, t = 10 minutes

Rate of reaction = $-\frac{\Delta [A]}{\Delta t}$

Change in concentration, $\Delta [A] = 0.4 - 0.5 = -0.1 \mathrm{mol\,L^{-1}}$

Rate = $-\frac{-0.1}{10}$ = 0.01 mol L⁻¹ min⁻¹

Since the reaction is $2A \rightarrow$ Products, rate of reaction = $\frac{1}{2} \times \text{rate of disappearance of A} = \frac{1}{2} \times 0.01 = 0.005 \mathrm{mol\,L^{-1}\,min^{-1}}$

Hence, the rate of reaction during this interval is 0.005 mol L⁻¹ min⁻¹. — The rate is calculated from the change in concentration of A divided by time, then adjusted by the stoichiometric coefficient since 2 moles of A react per reaction event.

Question 3

For a reaction, A + B → Product; the rate law is given by, r = k[A]^{1/2}[B]^2. What is the order of the reaction?

Accepted Answer

The order of a reaction is the sum of the powers of the concentration terms in the rate law.

Given rate law: r = k[A]^{1/2}[B]^2

Order with respect to A = 1/2
Order with respect to B = 2

Overall order = 1/2 + 2 = 2.5 (or 5/2)

Therefore, the order of the reaction is 2.5. — The order of reaction is determined by adding the exponents of the concentration terms in the rate law. Here, the exponent of [A] is 1/2 and that of [B] is 2. Adding these gives 2.5.

Question 4

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Accepted Answer

For a second order reaction, rate = k[X]^2.

If concentration of X is increased to three times, new concentration = 3[X].

New rate = k(3[X])^2 = k * 9[X]^2 = 9 × original rate.

Therefore, the rate of formation of Y will increase by 9 times. — Since the reaction is second order with respect to X, rate depends on the square of concentration of X. Tripling the concentration increases the rate by 3^2 = 9 times.

Question 5

A first order reaction has a rate constant $1.15 	imes 10^{-3} \, \mathrm{s}^{-1}$. How long will 5 g of this reactant take to reduce to 3 g?

Accepted Answer

Given:
Rate constant, k = 1.15 × 10⁻³ s⁻¹
Initial amount, [A]₀ = 5 g
Final amount, [A] = 3 g

For a first order reaction, the integrated rate law is:
ln([A]₀/[A]) = kt

Calculate time t:
ln(5/3) = (1.15 × 10⁻³) × t
ln(1.6667) = 1.15 × 10⁻³ × t
0.5108 = 1.15 × 10⁻³ × t

Therefore,
t = 0.5108 / (1.15 × 10⁻³) = 444.17 s

Convert seconds to minutes:
444.17 s ÷ 60 = 7.40 minutes

Hence, it will take approximately 7.4 minutes for 5 g of the reactant to reduce to 3 g. — Step 1: Use the first order integrated rate law ln([A]₀/[A]) = kt.
Step 2: Substitute given values and solve for time t.
Step 3: Calculate natural logarithm and divide by rate constant.
Step 4: Convert time from seconds to minutes for convenience.

Question 6

Time required to decompose $\mathrm{SO_2Cl_2}$ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Accepted Answer

Given:
Half-life, t₁/₂ = 60 minutes
Reaction is first order.

For a first order reaction, the half-life is related to rate constant k by:
t₁/₂ = \frac{0.693}{k}

Rearranging to find k:
k = \frac{0.693}{t_{1/2}} = \frac{0.693}{60} = 0.01155 	ext{ min}^{-1}

Hence, the rate constant k = 0.01155 min⁻¹. — Step 1: Use the first order half-life formula t₁/₂ = 0.693/k.
Step 2: Substitute the given half-life and solve for k.
Step 3: Calculate the numerical value of k.

Question 7

What will be the effect of temperature on rate constant?

Accepted Answer

The rate constant (k) of a chemical reaction increases with an increase in temperature. According to the Arrhenius equation, k = A e^(-Ea/RT), where Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. As temperature increases, the exponential term increases because -Ea/RT becomes less negative, leading to a higher rate constant. Thus, increasing temperature increases the rate constant and hence the rate of reaction. — From Arrhenius equation: k = A e^(-Ea/RT). As T increases, the value of -Ea/RT becomes less negative, so e^(-Ea/RT) increases, increasing k. Therefore, temperature has a direct effect on increasing the rate constant.

Question 8

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate $E_{\mathrm{a}}$.

Accepted Answer

Given: Rate doubles when temperature increases by 10 K from 298 K to 308 K.

Let k1 be the rate constant at T1 = 298 K, and k2 be the rate constant at T2 = 308 K.

Given: k2 = 2 k1

From Arrhenius equation:
$$
\ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)
$$

Substitute values:
$$
\ln 2 = \frac{E_a}{8.314} \left( \frac{1}{298} - \frac{1}{308} \right)
$$

Calculate the difference:
$$
\frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{91784} = 1.089 \times 10^{-4} K^{-1}
$$

Calculate $ \ln 2 = 0.693 $

Now,
$$
0.693 = \frac{E_a}{8.314} \times 1.089 \times 10^{-4}
$$

Rearranged,
$$
E_a = \frac{0.693 \times 8.314}{1.089 \times 10^{-4}} = \frac{5.758}{1.089 \times 10^{-4}} = 52860 \text{ J/mol} = 52.86 \text{ kJ/mol}
$$

Therefore, the activation energy, $ E_a $, is approximately 52.9 kJ/mol. — Using the Arrhenius equation ratio form and the given doubling of rate constant with 10 K increase, we calculate Ea stepwise as shown above.

Question 9

The activation energy for the reaction

$$
2 \mathrm{HI}(\mathrm{g}) \rightarrow \mathrm{H}_2 + \mathrm{I}_2(\mathrm{g})
$$

is 209.5 kJ mol⁻¹ at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Accepted Answer

Given:
Activation energy, Ea = 209.5 kJ/mol = 209500 J/mol
Temperature, T = 581 K
Gas constant, R = 8.314 J/mol·K

The fraction of molecules having energy equal to or greater than Ea is given by the Arrhenius factor:
$$
\text{Fraction} = e^{-E_a/(RT)}
$$

Calculate the exponent:
$$
\frac{E_a}{RT} = \frac{209500}{8.314 \times 581} = \frac{209500}{4829.234} \approx 43.38
$$

Therefore,
$$
\text{Fraction} = e^{-43.38} \approx 1.56 \times 10^{-19}
$$

This means only a very small fraction (~1.56 × 10⁻¹⁹) of molecules have energy equal to or greater than the activation energy at 581 K. — Using the Arrhenius equation fraction term, the exponential of negative Ea/RT gives the fraction of molecules with sufficient energy. Substituting values and calculating the exponent leads to the very small fraction as shown.

Question 10

1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3NO(g) → N2O(g) Rate = k[NO]^2
(ii) H2O2(aq) + 3I^-(aq) + 2H^+ → 2H2O(l) + I3^- Rate = k[H2O2][I^-]
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k[CH3CHO]^{3/2}
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k[C2H5Cl]

Accepted Answer

Solution:
(i) Rate = k[NO]^2
Order = 2 (since exponent of [NO] is 2)
Dimensions of k:
Rate has units mol L^{-1} s^{-1}
[NO] has units mol L^{-1}
So, k units = (mol L^{-1} s^{-1}) / (mol L^{-1})^2 = mol^{-1} L s^{-1}

(ii) Rate = k[H2O2][I^-]
Order = 1 + 1 = 2
Dimensions of k:
Rate units = mol L^{-1} s^{-1}
Concentration units = mol L^{-1}
So, k units = (mol L^{-1} s^{-1}) / (mol L^{-1} * mol L^{-1}) = mol^{-1} L s^{-1}

(iii) Rate = k[CH3CHO]^{3/2}
Order = 3/2 = 1.5
Dimensions of k:
Rate units = mol L^{-1} s^{-1}
Concentration units = mol L^{-1}
So, k units = (mol L^{-1} s^{-1}) / (mol L^{-1})^{3/2} = mol^{-0.5} L^{0.5} s^{-1}

(iv) Rate = k[C2H5Cl]
Order = 1
Dimensions of k:
Rate units = mol L^{-1} s^{-1}
Concentration units = mol L^{-1}
So, k units = (mol L^{-1} s^{-1}) / (mol L^{-1}) = s^{-1} — The order of reaction is determined by the sum of the exponents of concentration terms in the rate expression. The units of rate constant k are derived by balancing the units in the rate equation: Rate (mol L^{-1} s^{-1}) = k * [concentration terms].

Question 11

2 For the reaction:

2 A + B → A2B

the rate = k[A][B]^2 with k = 2.0 × 10^{-6} mol^{-2} L^2 s^{-1}. Calculate the initial rate of the reaction when [A] = 0.1 mol L^{-1}, [B] = 0.2 mol L^{-1}. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^{-1}.

Accepted Answer

Given:
k = 2.0 × 10^{-6} mol^{-2} L^2 s^{-1}
Initial concentrations: [A] = 0.1 mol L^{-1}, [B] = 0.2 mol L^{-1}

Rate = k [A][B]^2

Initial rate:
= 2.0 × 10^{-6} × 0.1 × (0.2)^2
= 2.0 × 10^{-6} × 0.1 × 0.04
= 8.0 × 10^{-9} mol L^{-1} s^{-1}

After [A] reduced to 0.06 mol L^{-1}:
Rate = 2.0 × 10^{-6} × 0.06 × (0.2)^2
= 2.0 × 10^{-6} × 0.06 × 0.04
= 4.8 × 10^{-9} mol L^{-1} s^{-1} — Substitute the given concentrations into the rate law and calculate the rate for both initial and changed concentrations.

Question 12

3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10^{-4} mol^{-1} L s^{-1}?

Accepted Answer

Given: Reaction is zero order in NH3.
Reaction: 2 NH3 → N2 + 3 H2
Rate of disappearance of NH3 = k = 2.5 × 10^{-4} mol L^{-1} s^{-1}

From stoichiometry:
Rate of formation of N2 = (1/2) × Rate of disappearance of NH3 = (1/2) × 2.5 × 10^{-4} = 1.25 × 10^{-4} mol L^{-1} s^{-1}

Rate of formation of H2 = (3/2) × Rate of disappearance of NH3 = (3/2) × 2.5 × 10^{-4} = 3.75 × 10^{-4} mol L^{-1} s^{-1} — Using stoichiometric coefficients, relate rate of disappearance of NH3 to rates of formation of N2 and H2. Since reaction is zero order, rate = k.

Question 13

4 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by

Rate = k [CH3OCH3]^{3/2}

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

Rate = k (p_{CH3OCH3})^{3/2}

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Accepted Answer

Given:
Rate = k (p)^{3/2}
Pressure unit = bar
Time unit = minutes

Units of rate:
Rate = change in pressure / time = bar min^{-1}

Units of k:
Rate = k × (pressure)^{3/2}
=> k = Rate / (pressure)^{3/2}
=> Units of k = (bar min^{-1}) / (bar)^{3/2} = bar^{1 - 3/2} min^{-1} = bar^{-1/2} min^{-1} — Rate is change in pressure per unit time, so units are bar min^{-1}. The rate constant units are derived by dividing rate units by pressure units raised to the power 3/2.

Question 14

10 In a reaction between A and B, the initial rate of reaction (r_0) was measured for different initial concentrations of A and B as given below:

| A/ mol L⁻¹ | 0.20 | 0.20 | 0.40 |
| B/ mol L⁻¹ | 0.30 | 0.10 | 0.05 |
| r₀/mol L⁻¹ s⁻¹ | 5.07 × 10⁻⁵ | 5.07 × 10⁻⁵ | 1.43 × 10⁻⁴ |

What is the order of the reaction with respect to A and B?

Accepted Answer

Let rate = k [A]^m [B]^n
From experiments 1 and 2:
[A] same (0.20), [B] changes from 0.30 to 0.10
Rate same (5.07 × 10^{-5})
Since rate unchanged when [B] changes, order with respect to B (n) = 0

From experiments 2 and 3:
[B] changes from 0.10 to 0.05 (halved), [A] changes from 0.20 to 0.40 (doubled)
Rate changes from 5.07 × 10^{-5} to 1.43 × 10^{-4}
Rate ratio = 1.43 × 10^{-4} / 5.07 × 10^{-5} ≈ 2.82

Since n=0, rate ratio = (2)^m
2.82 ≈ 2^m
Taking log:
log 2.82 = m log 2
0.45 = m × 0.301
m ≈ 1.5

Therefore, order with respect to A is approximately 1.5, and order with respect to B is 0. — By comparing rates at different concentrations and keeping one reactant concentration constant, the order with respect to each reactant is determined.

Question 15

11 The following results have been obtained during the kinetic studies of the reaction:

2 A + B → C + D

| Experiment | [A]/mol L⁻¹ | [B]/mol L⁻¹ | Initial rate of formation of D/mol L⁻¹ min⁻¹ |
| I | 0.1 | 0.1 | 6.0 × 10⁻³ |
| II | 0.3 | 0.2 | 7.2 × 10⁻² |
| III | 0.3 | 0.4 | 2.88 × 10⁻¹ |
| IV | 0.4 | 0.1 | 2.40 × 10⁻² |

Determine the rate law and the rate constant for the reaction.

Accepted Answer

Let rate = k [A]^m [B]^n
Using experiments I and II:
Rate ratio = (7.2 × 10^{-2}) / (6.0 × 10^{-3}) = 12
[A] ratio = 0.3 / 0.1 = 3
[B] ratio = 0.2 / 0.1 = 2

So, 12 = 3^m × 2^n

Using experiments II and III:
Rate ratio = (2.88 × 10^{-1}) / (7.2 × 10^{-2}) = 4
[A] same (0.3)
[B] ratio = 0.4 / 0.2 = 2

So, 4 = 2^n => n = 2

Using experiments I and IV:
Rate ratio = (2.40 × 10^{-2}) / (6.0 × 10^{-3}) = 4
[A] ratio = 0.4 / 0.1 = 4
[B] ratio = 0.1 / 0.1 = 1

So, 4 = 4^m => m = 1

Therefore, rate law: rate = k [A]^1 [B]^2

Calculate k using experiment I:
6.0 × 10^{-3} = k × 0.1 × (0.1)^2 = k × 0.001
k = 6.0 mol^{-2} L^{2} min^{-1} — By comparing rates and concentrations from different experiments, the orders m and n are found. Then k is calculated by substituting values from any experiment.

Chemical Kinetics

Chemical Kinetics — Study Notes

Chemical Kinetics

3.1 Rate of a Chemical Reaction

3.2 Factors Influencing Rate of a Reaction

Practice Questions — Chemical Kinetics

All 5 Chapters in Chemistry-I