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:

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

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

  5. 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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