Question 5 of 37intermediate🔧 ApplyNumerical2 marks

A first order reaction has a rate constant 1.15×10−3 s−11.15 \times 10^{-3} \text{ s}^{-1}. How long will 5 g of this reactant take to reduce to 3 g?

Correct Answer

444 s

Exercise: Intext Questions | Q: 3.5 | (Page 18)
For More Understanding

Explanation

This question applies the integrated rate law for a first-order reaction provided in the context. Since the reaction is first order, the rate constant unit is s⁻¹, and the ratio of masses can be used directly in place of concentration ratio because volume cancels out. The formula used is derived from the relationship k=2.303tlog⁡[R]0[R]k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} found in the 'Half-Life of a Reaction' section.

Solution Steps

  1. Step 1: Identify the formula For a first order reaction, the integrated rate law is given by:

    t=2.303klog⁡[R]0[R]t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

  2. Step 2: Substitute the values Given: Rate constant k=1.15×10−3 s−1k = 1.15 \times 10^{-3} \text{ s}^{-1} Initial amount [R]0=5 g[R]_0 = 5 \text{ g} Final amount [R]=3 g[R] = 3 \text{ g}

    t=2.3031.15×10−3log⁡(53)t = \frac{2.303}{1.15 \times 10^{-3}} \log \left( \frac{5}{3} \right)

  3. Step 3: Calculate the result

    t=2.3031.15×10−3log⁡(1.667)t = \frac{2.303}{1.15 \times 10^{-3}} \log (1.667) t=2002.6×0.2218t = 2002.6 \times 0.2218 t=444.38 st = 444.38 \text{ s}

    Rounding off, the time taken is 444 s.

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