Question

Calculate the rate constant of the first order reaction if 80% of the reactant decomposes in 60 minutes.

• A. $2.68 \times 10^{-2}\ \text{minute}^{-1}$
• B. $5.36 \times 10^{-2}\ \text{minute}^{-1}$
• C. $1.34 \times 10^{-2}\ \text{minute}^{-1}$
• D. $8.1 \times 10^{-2}\ \text{minute}^{-1}$

Hint:

For first order reactions, if percentage decomposed is given, first calculate fraction remaining and use: \[ k=\frac{2.303}{t}\log\frac{\text{Initial{\text{Remaining \]

Answer 1

The Correct Option is A

Step 1: State the first order rate equation. For a first order reaction: \[ k=\frac{2.303}{t}\log\frac{a}{a-x} \] where:

• $k$ = rate constant
• $t$ = time
• $a$ = initial concentration
• $x$ = amount decomposed

Step 2: Determine amount remaining after decomposition. Given $80%$ reactant decomposes in $60$ minutes. Therefore, reactant left: \[ 100%-80%=20% \] So, \[ \frac{a-x}{a}=0.20 \] Hence: \[ \frac{a}{a-x}=\frac{1}{0.20}=5 \]
Step 3: Substitute values in formula. \[ k=\frac{2.303}{60}\log 5 \] Using: \[ \log 5 = 0.6990 \] \[ k=\frac{2.303 \times 0.6990}{60} \]
Step 4: Calculate rate constant. \[ k=\frac{1.6098}{60} \] \[ k=0.0268\ \text{minute}^{-1} \] \[ k=2.68\times10^{-2}\ \text{minute}^{-1} \]
Step 5: Conclude the correct option. Therefore, the rate constant is: \[ \boxed{2.68\times10^{-2}\ \text{minute}^{-1 \]