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(B) For a general chemical reaction $aA + bB 	o cC + dD$, the rate of reaction is expressed as: $	ext{rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{

Vedclass · Accepted Answer

$6A + 4B 	o 3C + 4D$

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(B) For a general chemical reaction $aA + bB 	o cC + dD$, the rate of reaction is expressed as: $	ext{rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$. Comparing the given rate expression with the general form: $-\frac{1}{6} \frac{d[A]}{dt} = -\frac{1}{a} \frac{d[A]}{dt} \implies a = 6$. $-\frac{1}{4} \frac{d[B]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} \implies b = 4$. $\frac{1}{3} \frac{d[C]}{dt} = \frac{1}{c} \frac{d[C]}{dt} \implies c = 3$. $\frac{1}{4} \frac{d[D]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \implies d = 4$. Substituting these stoichiometric coefficients into the general reaction equation, we get: $6A + 4B 	o 3C + 4D$.

Select the correct reaction for the given rate: $\text{rate} = -\frac{1}{6} \frac{d[A]}{dt} = -\frac{1}{4} \frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} = \frac{1}{4} \frac{d[D]}{dt}$

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