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Question: A gaseous hypothetical chemical equation \(2A\)⇌\(4B + C\)is carried out in a closed vessel. The con...

A gaseous hypothetical chemical equation 2A2A⇌4B+C4B + Cis carried out in a closed vessel. The concentration of B is found to increase by 5×10−3moll−15 \times 10^{- 3}moll^{- 1} in 10 second. The rate of appearance of B is

A

5×10−4moll−1sec⁡−15 \times 10^{- 4}moll^{- 1}\sec^{- 1}{}

B

5×10−5moll−1sec⁡−15 \times 10^{- 5}moll^{- 1}\sec^{- 1}{}

C

6×10−5moll−1sec⁡−16 \times 10^{- 5}moll^{- 1}\sec^{- 1}{}

D

4×10−4moll−1sec⁡−14 \times 10^{- 4}moll^{- 1}\sec^{- 1}{}

Answer

5×10−4moll−1sec⁡−15 \times 10^{- 4}moll^{- 1}\sec^{- 1}{}

Explanation

Solution

Increase in concentration of B=5×10−3moll−1B = 5 \times 10^{- 3}moll^{- 1},

Time = 10 sec.

Rate of appearance of B=Increase of concentration of ∫=5×10−3moll−110sec−4−1sec⁡−1= \frac{\text{Increase of concentration of }\int_{}^{}}{} = \frac{5 \times 10^{- 3}moll^{- 1}}{10se{c^{- 4}}^{- 1}\sec^{- 1}}.