Question 24 of 37intermediate🔧 ApplyNumerical2 marks

The rate constant for a first order reaction is 60 s−160 \text{ s}^{-1}. How much time will it take to reduce the initial concentration of the reactant to its 1/16th1/16^{\text{th}} value?

Correct Answer

For a first order reaction, the integrated rate equation is:

k = (2.303/t) log([R]₀/[R])

Given: k = 60 s⁻¹ and [R] = [R]₀/16

Substituting the values: 60 = (2.303/t) log([R]₀/([R]₀/16)) 60 = (2.303/t) log(16)

Since log(16) = log(2⁴) = 4×0.3014 \times 0.301 = 1.204

60 = (2.303/t) × 1.204

t = (2.303×1.2042.303 \times 1.204)/60 t = 2.773/60 = 0.0462 s

Therefore, the time required is 4.62×104.62 \times 10⁻² s.

Exercise: EXERCISES | Q: 3.16 | (Page 27)
For More Understanding

Explanation

This question tests understanding of the integrated rate law for first order reactions. The key formula from the textbook is k = (2.303/t) log([R]₀/[R]). When concentration reduces to 1/16th, the ratio [R]₀/[R] = 16. Using log(16) = 1.204, we can solve for time t. The calculation shows that approximately 0.046 seconds are needed for the concentration to drop to 1/16th of its initial value.

Solution Steps

  1. Step 1: Write the integrated rate equation for first order reaction: k = (2.303/t) log([R]₀/[R])

  2. Step 2: Substitute [R] = [R]₀/16, so [R]₀/[R] = 16

  3. Step 3: Calculate log(16) = log(2⁴) = 4×0.3014 \times 0.301 = 1.204

  4. Step 4: Substitute k = 60 s⁻¹ and solve: 60 = (2.303/t) × 1.204

  5. Step 5: Calculate t = (2.303×1.2042.303 \times 1.204)/60 = 2.773/60 = 0.0462 s

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