JEE Chemistry 2023 Question with Solution

Given: The reaction is $A + B \rightarrow \text{Product}$ and the order is first order with respect to both $A$ and $B$.

Find: The values of $x$ and $Y$.

The rate law is

$$r = k[A][B]$$

Using the given data:

$$0.10 = k(20)(0.5)$$ $$0.40 = k(x)(0.5)$$ $$0.80 = k(40)(Y)$$

For $x$, divide the second equation by the first:

$$\frac{0.40}{0.10} = \frac{k(x)(0.5)}{k(20)(0.5)}$$ $$4 = \frac{x}{20}$$ $$x = 80$$

For $Y$, divide the third equation by the first:

$$\frac{0.80}{0.10} = \frac{k(40)(Y)}{k(20)(0.5)}$$ $$8 = \frac{40Y}{10}$$ $$80 = 40Y$$ $$Y = 2$$

Therefore, the values of $x$ and $Y$ are $80$ and $2$, respectively. The correct option from the working is A. The solution labels option D, but that conflicts with the extracted working and option list.

Given: Rate law is first order in $A$ and first order in $B$.

Find: Unknown concentrations $x$ and $Y$.

Since

$$r \propto [A][B]$$

keeping $[B]$ constant at $0.5$, the rate is directly proportional to $[A]$.

From the first and second rows:

$$\frac{0.40}{0.10} = 4$$

So $[A]$ must also become $4$ times:

$$x = 4 \times 20 = 80$$

Now compare the first and third rows:

$$\frac{0.80}{0.10} = 8$$

But $[A]$ changes from $20$ to $40$, which is only a factor of $2$. Therefore $[B]$ must contribute the remaining factor of $4$:

$$2 \times \frac{Y}{0.5} = 8$$ $$\frac{Y}{0.5} = 4$$ $$Y = 2$$

Hence, $x = 80$ and $Y = 2$.