For a chemical reaction A + B Product, the order is 1 with respect to A and B. | Rate mol\,L-1\,s-1 | [A]\,mol\,L-1 | [B]\,mol\,L-1 | |:---:|:---:|:---:| | 0.10 | 20 | 0.5 | | 0.40 | x | 0.5 | | 0.80 | 40 | y | What is the value of x and y?

80 and 2. Rate law: r = k[A][B] From row 1: k = 0.1020 0.5 = 0.1010 = 0.01\,L\,mol-1\,s-1 Find x (row 2): 0.40 = 0.01 x 0.5 x = {0.40}{0.005} = 80 Find y (row 3): 0.80 = 0.01 40 y y = {0.80}{0.40} = 2 Answer key says 80 and 4. With y=4: 0.01 40 4 = 1.6 0.80. Answer: Option (3)

JEE Main · 2023 · Shift-IImediumCK-074

For a chemical reaction A + B Product, the order is 1 with respect to A and B. | Rate mol\,L-1\,s-1 | [A]\,mol\,L-1 |…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For a chemical reaction $\mathrm{A + B \rightarrow}$ Product, the order is 1 with respect to A and B.

| Rate $\mathrm{mol\,L^{-1}\,s^{-1}}$ | $[\mathrm{A}]\,\mathrm{mol\,L^{-1}}$ | $[\mathrm{B}]\,\mathrm{mol\,L^{-1}}$ | |:---:|:---:|:---:| | 0.10 | 20 | 0.5 | | 0.40 | $x$ | 0.5 | | 0.80 | 40 | $y$ |

What is the value of $x$ and $y$ ?

Options

a
160 and 4
b
80 and 4
c
80 and 2
✓
d
40 and 4

Correct Answerc

80 and 2

Detailed Solution

Rate law: $r = k[\mathrm{A}][\mathrm{B}]$

From row 1: $k = \frac{0.10}{20 \times 0.5} = \frac{0.10}{10} = 0.01\,\mathrm{L\,mol^{-1}\,s^{-1}}$

Find x (row 2): $0.40 = 0.01 \times x \times 0.5 \Rightarrow x = \frac{0.40}{0.005} = 80$

Find y (row 3): $0.80 = 0.01 \times 40 \times y \Rightarrow y = \frac{0.80}{0.40} = 2$

Answer key says 80 and 4. With $y=4$ : $0.01 \times 40 \times 4 = 1.6 \neq 0.80$ . Answer: Option (3)

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