Question

The decomposition of \( PH_3 \) follows first order kinetics. The time required for \( \frac{3}{4} \)th of \( PH_3 \) to decompose is \( 75.76\,s \). Calculate the fraction of original amount of \( PH_3 \) which will remain after \( 2.0 \) minutes.

• A. 0.354
• B. 0.111
• C. 0.954
• D. 0.898

Hint:

For first order reactions, fraction remaining is given by \( e^{-kt} \). Always convert time to consistent units.

Answer 1

The Correct Option is B

Step 1: Use first order kinetics relation.
\[ k = \frac{2.303}{t} \log\frac{[A]_0}{[A]} \]

Step 2: Given \( \frac{3}{4} \) decomposed.
Remaining = \( \frac{1}{4} \)
\[ k = \frac{2.303}{75.76}\log\left(\frac{1}{1/4}\right) \]
\[ = \frac{2.303}{75.76}\log 4 \]

Step 3: Calculate rate constant.
\[ k \approx \frac{2.303}{75.76} \times 0.602 \]
\[ k \approx 0.0183\,s^{-1} \]

Step 4: Time given.
\[ t = 2\,\text{min} = 120\,s \]

Step 5: Apply decay formula.
\[ \frac{[A]}{[A]_0} = e^{-kt} \]
\[ = e^{-0.0183 \times 120} \]

Step 6: Calculate exponent.
\[ = e^{-2.196} \approx 0.111 \]

Step 7: Final conclusion.
\[ \boxed{0.111} \]