In the chemical reaction, 3A + B > 2C + 3D, the rate of appearance of C is reported as 1 mol L^1 s^1. What is the rate of disappearance of A?
Question
In the chemical reaction, $3A + B \rightarrow 2C + 3D$, the rate of appearance of C is reported as 1 mol L$^{1}$ s$^{1}$.What is the rate of disappearance of A?
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Answer
To determine the rate of disappearance of ( A ) in the reaction ( 3A + B \rightarrow 2C + 3D ), given that the rate of appearance of ( C ) is ( 1 , \text{mol L}^{1} \text{s}^{1} ), follow these steps:

Identify the given information:
 Reaction: ( 3A + B \rightarrow 2C + 3D )
 Rate of appearance of ( C ): ( \frac{d[C]}{dt} = 1 , \text{mol L}^{1} \text{s}^{1} )

Recall the relationships between the rates of disappearance or appearance of reactants and products using the stoichiometric coefficients:
 For ( A ): ( \frac{1}{3} \frac{d[A]}{dt} )
 For ( C ): ( \frac{1}{2} \frac{d[C]}{dt} )

Express the rate of the reaction in terms of the rate of disappearance of ( A ) and appearance of ( C ): $$ \frac{1}{3} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[C]}{dt} $$

Substitute ( \frac{d[C]}{dt} ) with the given value: $$ \frac{1}{3} \frac{d[A]}{dt} = \frac{1}{2} \times 1 , \text{mol L}^{1} \text{s}^{1} $$

Solve for ( \frac{d[A]}{dt} ): $$ \frac{1}{3} \frac{d[A]}{dt} = 0.5 , \text{mol L}^{1} \text{s}^{1} $$ $$ \frac{d[A]}{dt} = 3 \times 0.5 , \text{mol L}^{1} \text{s}^{1} $$ $$ \frac{d[A]}{dt} = 1.5 , \text{mol L}^{1} \text{s}^{1} $$
So, the rate of disappearance of ( A ) is 1.5 mol ( \text{L}^{1} \text{s}^{1} ).
Final Answer: ( 1.5 , \text{mol L}^{1} \text{s}^{1} )
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