AAO SIKHTE HAI

🔹 Exercise: Questions 4.1–4.5 (Page 117–118)

Q4.1
Find the order and dimensions of rate constants for these rate laws:
a) Rate = k [NO]² → Order = 2, k has units M⁻¹ s⁻¹
b) Rate = k [H₂O₂][I⁻] → Order = 2, k has units M⁻¹ s⁻¹
c) Rate = k [CH₃CHO]³⁄² → Order = 1.5, units of k: M⁻½ s⁻¹
d) Rate = k [C₂H₅Cl] → Order = 1, units M⁰ s⁻¹ = s⁻¹
Q4.2
For 2A + B → A₂B, Rate = k[A][B]² (k = 2×10⁻⁶ M⁻² s⁻¹). Calculate initial rate with [A]=0.1, [B]=0.2 and then with [A]=0.06, assuming B drops to 0.18.
- Initial: Rate = 2×10⁻⁶ ×0.1×(0.2)² = 8.0×10⁻⁹ M s⁻¹
- After change: Rate ≈ 3.89×10⁻⁹ M s⁻¹
Q4.3
Decomposition of NH₃ on Pt (zero-order, k=2.5×10⁻⁴ M s⁻¹). Find rates of N₂ and H₂ formation.
For each mole NH₃ decomposed: produces ½ N₂ and 3/2 H₂ →
- Rate(N₂) = ½ k = 1.25×10⁻⁴ M s⁻¹
- Rate(H₂) = (3/2) k = 3.75×10⁻⁴ M s⁻¹
Q4.4
Dimethyl ether decomposition, Rate = k [CH₃OCH₃]³⁄². Units if pressure in bar and time in min?
- Unit of rate: bar min⁻¹
- k has units = bar^(-½) min⁻¹
Q4.5
List factors that affect chemical reaction rate.

Concentration (or pressure for gases)
Temperature
Presence of catalyst

🔹 Exercise Questions and Answers (Q.4.6 to Q.4.15)

Q4.6

A reaction is first-order in A. After 15 minutes, 75% of A has reacted. What is the half-life?

Answer:
75% reaction → 25% remains

\text{Use: } k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) = \frac{2.303}{15} \log\left(\frac{100}{25}\right)

k = \frac{2.303}{15} \log 4 = \frac{2.303 \times 0.602}{15} ≈ 0.0925\ \text{min}^{-1}

t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0925} ≈ 7.49\ \text{min}

Q4.7

What will be the effect of temperature on rate constant?

Answer:
According to the Arrhenius equation:

k = A e^{-E_a/RT}

An increase in temperature increases k exponentially, thereby increasing the reaction rate.

Q4.8

Why is the rate constant k independent of initial concentrations?

Answer:
Because k depends only on temperature and activation energy, not on concentrations. The rate, however, does depend on concentrations.

Q4.9

A reaction is first-order in A. Time for 60% completion is 10 minutes. What is the time for 90% completion?

Answer:
For 60% completion:

[A]_0 = 100,\ [A] = 40 \Rightarrow k = \frac{2.303}{10} \log\left(\frac{100}{40}\right) ≈ 0.0920\ \text{min}^{-1}

Now for 90% completion:

[A] = 10,\ k = 0.0920 t = \frac{2.303}{k} \log\left(\frac{100}{10}\right) = \frac{2.303}{0.0920} \log 10 = \frac{2.303}{0.0920} ≈ 25.03\ \text{min}

Q4.10

Plot log₁₀[R] vs time for a first-order reaction. What does the slope represent?

Answer:
A straight-line graph is obtained.
Slope = –k/2.303, where k is the rate constant.

Q4.11

The rate constant for a first-order reaction is 60 s⁻¹. How much time will it take to reduce the reactant to 1/16th of its initial value?

Answer:

\frac{[R]_0}{[R]} = 16 \Rightarrow \log(16) = 1.204 t = \frac{2.303}{k} \log(16) = \frac{2.303 \times 1.204}{60} ≈ 0.0462\ \text{s}

Q4.12

Write the Arrhenius equation. What does each symbol mean?

Answer:
k = A e^(-Eₐ/RT)
Where:

k = rate constant
A = frequency factor
Eₐ = activation energy
R = gas constant
T = temperature (K)

Q4.13

Arrhenius plot of log k vs 1/T gives a straight line. What does the slope represent?

Answer:
Slope = –Ea / 2.303R
From log k = log A – Ea / (2.303RT)

Q4.14

The rate of a reaction doubles for every 10°C rise in temperature. Explain.

Answer:
According to the Arrhenius equation, rate constant increases exponentially with temperature, approximately doubles for every 10°C rise due to more molecules having energy ≥ Ea.

Q4.15

The activation energy of a reaction is 75.2 kJ/mol. Calculate rate constant at 298 K if A = 5 × 10⁹ s⁻¹.

Answer:
Use:

k = A e^{-E_a / RT}

E_a = 75.2 × 10³ J/mol,\ R = 8.314\ J/mol·K,\ T = 298\ K \Rightarrow \frac{E_a}{RT} = \frac{75200}{8.314 × 298} ≈ 30.29

k = 5 × 10⁹ × e^{-30.29} ≈ 5 × 10⁹ × 6.98 × 10^{-14} ≈ 3.49 × 10^{-4} s⁻¹

Monday, 7 July 2025