Question

The rate constants k₁ and k₂ for two different reactions are 10¹⁶ × e^-2000/T and 10¹⁵ × e^-1000/T respectively. The temperature at which k₁ = k₂ is :

• A. \(\frac{2000}{2.303}K\)
• B. 2000 K
• C. \(\frac{1000}{2.303}K\)
• D. 1000 K

Answer 1

The Correct Option is C

During the electrolysis of brine (aqueous NaCl solution) using inert electrodes, the ions present are:

• At Cathode (Reduction): Na⁺, H₂O molecules
• At Anode (Oxidation): Cl⁻, H₂O molecules

Step-by-Step Explanation:

At Cathode (negative electrode):

Reduction potentials:

• Na⁺(aq) + e⁻ → Na(s) (Highly negative potential)
• 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) (Lower negative potential)

H₂ gas is liberated at cathode since reduction of water is more favorable than Na⁺ reduction.

At Anode (positive electrode):

Oxidation potentials:

• 2Cl⁻(aq) → Cl₂(g) + 2e⁻ (Favorable oxidation)
• 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ (Less favorable than chloride oxidation in concentrated solutions)

Cl₂ gas is liberated at anode, as oxidation of chloride ions is preferred due to the higher concentration and easier oxidation potential.

Overall reaction for brine electrolysis:

2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)

Conclusion:

The correct option is: (D) Cl₂ liberates at anode

Correct Answer: Option (D)

Answer 2

The rate constant \( k \) for the reactions is given by the Arrhenius equation: 
\[ k = A \times e^{-\frac{E_a}{RT}} \] Where:
\( A \) is the pre-exponential factor,
\( E_a \) is the activation energy,
\( R \) is the gas constant (8.314 J/mol·K),
\( T \) is the temperature in Kelvin.

For the two reactions, we have the following rate constants: \[ k_1 = 10^{16} \times e^{-\frac{2000}{T}} \quad \text{and} \quad k_2 = 10^{15} \times e^{-\frac{1000}{T}} \] At the temperature when \( k_1 = k_2 \), we can set the two rate constants equal to each other: \[ 10^{16} \times e^{-\frac{2000}{T}} = 10^{15} \times e^{-\frac{1000}{T}} \] Dividing both sides by \( 10^{15} \), we get: \[ 10 \times e^{-\frac{2000}{T}} = e^{-\frac{1000}{T}} \] Taking the natural logarithm (ln) of both sides: \[ \ln(10) + \left( -\frac{2000}{T} \right) = -\frac{1000}{T} \] Simplifying: \[ \ln(10) = \frac{1000}{T} \] Solving for \( T \): \[ T = \frac{1000}{\ln(10)} = \frac{1000}{2.303} \] Therefore, the temperature at which \( k_1 = k_2 \) is \( \frac{1000}{2.303} \) K. 

Hence, the correct answer is (C) \( \frac{1000}{2.303} \) K.