Question: 83: For a chemical reaction
4 A+3 B \rightarrow 6 C+9 Drate of formation of C is 6 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and rate of disappearance of A is 4 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.
The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be :
(1) 1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
(2) 10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
(3) 1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
(4) 10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
Answer: Option (1)
Explanation:
The rate of reaction is defined using stoichiometric coefficients as:
\text{Rate} = \frac{1}{\nu_i} \left|\frac{d[\text{species}]}{dt}\right|For the given reaction, rate of formation of C is related as:
\text{Rate} = \frac{1}{6} \frac{d[C]}{dt}Substituting the given value:
\text{Rate} = \frac{1}{6} \times 6 \times 10^{-2} \text{Rate} = 1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}This matches the rate calculated using disappearance of A:
\text{Rate} = \frac{1}{4} \times 4 \times 10^{-2} = 1 \times 10^{-2}Thus, the rate of reaction is 1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.
Now, for consumption of B:
\text{Rate} = \frac{1}{3} \frac{d[B]}{dt} \frac{d[B]}{dt} = 3 \times \text{Rate} = 3 \times 1 \times 10^{-2} \frac{d[B]}{dt} = 3 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}Amount of B consumed in 10 seconds:
\Delta [B] = 3 \times 10^{-2} \times 10 \Delta [B] = 30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}Hence, the correct answer is Option (1).
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