Question

For $xA + yB$ $\to $ $zC$,
\[-\dfrac{d[A]}{dt}=-\dfrac{d[B]}{dt}=1.5\dfrac{d[C]}{dt}\]
then x, y and z are.
A. 1, 1, 1
B. 3, 2, 3
C. 3, 3, 2
D. 2, 2, 3

Answer 1

Hint: We should know how to write the change in concentration of the reactants versus change in concentration of the product in a chemical reaction.
For $xA + yB$ $\to $ $zC$ chemical reaction, the change in concentration of the reactants versus change in concentration of the product is as follows.
\[-\dfrac{1}{x}\dfrac{d[A]}{dt}=-\dfrac{1}{y}\dfrac{d[B]}{dt}=\dfrac{1}{z}\dfrac{d[C]}{dt}\]

Complete step by step solution:
- In the question it is asked to find the x, y and z values for the given chemical reaction.
- For xA + yB $\to $ zC chemical reaction, the change in concentration of the reactants versus change in the concentration of the products is as follows.
 \[-\dfrac{1}{x}\dfrac{d[A]}{dt}=-\dfrac{1}{y}\dfrac{d[B]}{dt}=\dfrac{1}{z}\dfrac{d[C]}{dt}\to (1)\]
- The given relationship between the reactants and the products in the question is as follows.
\[-\dfrac{d[A]}{dt}=-\dfrac{d[B]}{dt}=1.5\dfrac{d[C]}{dt}\]
- We can simplify the above relationship to get the values of the x, y and z.
\[-\dfrac{d[A]}{dt}=-\dfrac{d[B]}{dt}=1.5\dfrac{d[C]}{dt}\]
- Initially multiply the above equation with $\dfrac{1}{3}$
\[\begin{align}
  & -\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{3}\dfrac{d[B]}{dt}=\dfrac{1}{3}\times \dfrac{3}{2}\dfrac{d[C]}{dt} \\
 & -\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{3}\dfrac{d[B]}{dt}=\dfrac{1}{2}\dfrac{d[C]}{dt}\to (2) \\
\end{align}\]
- Now compare the equations (1) and (2) then we will get the x, y and z values.
- The x, y and z values are 3, 3, 2.
- Means 3moles of x reacts with 3 moles of y and forms 2 moles of the z as the product.

So, the correct option is C.

Note: We are supposed to multiply the given equation with a proper number to get the values of the x, y and Z. If we choose an improper number then it is very difficult to get the values of x, y and z form the given relationship.