For the reaction,A + 3B longrightarrow 2C,the | vedclass

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(C) For the reaction,$A + 3B \longrightarrow 2C$,the rate of reaction is given by: $-\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{2} \fra

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$2.8 \ mol \ dm^{-3} \ s^{-1}$

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(C) For the reaction,$A + 3B \longrightarrow 2C$,the rate of reaction is given by: $-\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$ Given that the rate of consumption of $A$ is $-\frac{d[A]}{dt} = 1.4 \ mol \ dm^{-3} \ s^{-1}$. Equating the terms for $A$ and $C$: $-\frac{d[A]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$ $1.4 = \frac{1}{2} \frac{d[C]}{dt}$ $\frac{d[C]}{dt} = 2 	imes 1.4 \ mol \ dm^{-3} \ s^{-1} = 2.8 \ mol \ dm^{-3} \ s^{-1}$

For the reaction,$A + 3B \longrightarrow 2C$,the rate of consumption of $A$ is $1.4 \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate of formation of $C$.

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Fill in the blanks:
$(a)$ The rate of reaction ....... as the concentration of reactant decreases.
$(b)$ The rate of reaction ....... as the concentration of product increases.

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