SATHEE: Chemical Kinetics Question 241

Chemical Kinetics Question 241

Question: A gaseous hypothetical chemical equation $2 A$ ⇌ $4 B + C$ is carried out in a closed vessel. The concentration of B is found to increase by $5 \times 10^{- 3} m o l l^{- 1}$ in 10 second. The rate of appearance of B is [AFMC 2001]

Options:

A) $5 \times 10^{- 4} m o l l^{- 1} s e c^{- 1}$

B) $5 \times 10^{- 5} m o l l^{- 1} s e c^{- 1}$

C) $6 \times 10^{- 5} m o l l^{- 1} \sec^{- 1}$

D) $4 \times 10^{- 4} m o l, l^{- 1} \sec^{- 1}$

Show Answer

Answer:

Correct Answer: A

Solution:

Increase in concentration of $B = 5 \times 10^{- 3} m o l l^{- 1}$

$T i m e = 10 \sec$ Rate of appearance of $B =$

$\frac{I n c r e a s e, o f, c o n c . B}{T i m e t a k e n}$

$= \frac{5 \times 10^{- 3} m o l l^{- 1}}{10 \sec} = 5 \times 10^{- 4} m o l l^{- 1} S e c^{- 1}$

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