Question

For a chemical reaction A + B —q Product. the order is 1 with respect to A and B. What is the value of x and y?

Rate mol L-1S-1	[A] mol L-1	[B] mol L-1
0.10	20	0.5
0.40	x	0.5
0.80	40	y

• A. 80 and 4
• B. 160 and 4
• C. 40 and 4
• D. 80 and 2

Answer 1

The Correct Option is D

Given: The reaction \( A + B \to \text{Product} \) is first order with respect to both \( A \) and \( B \). The rate law is:

\[ r = k[A][B] \]

where:

• \( r \): Rate of the reaction
• \( k \): Rate constant
• \( [A] \) and \( [B] \): Concentrations of \( A \) and \( B \), respectively

Step 1: Set Up Equations

Using the given data, we can write the following equations:

1. \( 0.10 = k(20)(0.5) \) — (1)
2. \( 0.40 = k(x)(0.5) \) — (2)
3. \( 0.80 = k(40)(Y) \) — (3)

Step 2: Solve for \( x \)

Divide equation (2) by equation (1):

\[ \frac{0.40}{0.10} = \frac{k(x)(0.5)}{k(20)(0.5)} \]

Simplify:

\[ 4 = \frac{x}{20} \quad \Rightarrow \quad x = 4 \times 20 = 80 \]

Step 3: Solve for \( Y \)

Divide equation (3) by equation (1):

\[ \frac{0.80}{0.10} = \frac{k(40)(Y)}{k(20)(0.5)} \]

Simplify:

\[ 8 = \frac{40Y}{10} \quad \Rightarrow \quad 80 = 40Y \quad \Rightarrow \quad Y = \frac{80}{40} = 2 \]

Final Answer:

The values of \( x \) and \( Y \) are 80 and 2, respectively.