Q17. In a reaction A and B reacts to form product. The
initial rate of reaction (ro) was determined

Q: What is the correct answer for CUET UG Chemistry Q17?

The correct answer is a. Assume rate law: r₀ = k[A]ˣ[B]ʸComparing Exp 1 and Exp 2 (keeping [A] constant):(6.81×10⁻⁴) / (2.27×10⁻⁴) = (0.30/0.10)ʸ ⇒ 3 = 3ʸ ⇒ y = 1.Comparing Exp 3 and Exp 1 (keeping [B] constant):(13.62×10⁻⁴) / (6.81×10⁻⁴) = (0.20/0.10)ˣ ⇒ 2 = 2ˣ ⇒ x = 1.So, the rate law is r₀ = k[A][B].Using Exp 2 to find k: 2.27×10⁻⁴ = k[0.10][0.10] = k(0.01) ⇒ k = 2.27×10⁻² L mol⁻¹ s⁻¹.Now, calculate r₀ for [A]=0.50 mol

Q: Q17. In a reaction A and B reacts to form product. The initial rate of reaction (ro) was determined using different initial concentration of A and B. [Table with experimental data] A/mol L-1: 0.10,…

The correct answer is a. Assume rate law: r₀ = k[A]ˣ[B]ʸComparing Exp 1 and Exp 2 (keeping [A] constant):(6.81×10⁻⁴) / (2.27×10⁻⁴) = (0.30/0.10)ʸ ⇒ 3 = 3ʸ ⇒ y = 1.Comparing Exp 3 and Exp 1 (keeping [B] constant):(13.62×10⁻⁴) / (6.81×10⁻⁴) = (0.20/0.10)ˣ ⇒ 2 = 2ˣ ⇒ x = 1.So, the rate law is r₀ = k[A][B].Using Exp 2 to find k: 2.27×10⁻⁴ = k[0.10][0.10] = k(0.01) ⇒ k = 2.27×10⁻² L mol⁻¹ s⁻¹.Now, calculate r₀ for [A]=0.50 mol

Q: What are the options for this CUET UG Chemistry question?

A. 56.75 x 10-4 mol L-1s-1 · B. 5.675 x 10-4 mol L-1s-1 · C. 21.67 x 10-4 mol L-1s-1 · D. 2.167 x 10-4 mol L-1s-1

Question

Q17. In a reaction A and B reacts to form product. The
initial rate of reaction (ro) was determined using
different initial concentration of A and B.
[Table with experimental data]
A/mol L-1: 0.10, 0.10, 0.20
B/mol L-1: 0.30, 0.10, 0.30
Ro/mol L-1 s-1: 6.81×10-4, 2.27×10-4, 13.62×10-4
What is the initial rate of reaction (ro) when critical
concentration of A and B is 0.50 mol/L and 0.50 mol/L
respectively.
(a) 56.75 x 10-4 mol L-1s-1
(b) 5.675 x 10-4 mol L-1s-1
(c) 21.67 x 10-4 mol L-1s-1
(d) 2. 167 x 10-4 mol L-1s-1

FirstInTest · Accepted Answer

Assume rate law: r₀ = k[A]ˣ[B]ʸComparing Exp 1 and Exp 2 (keeping [A] constant):(6.81×10⁻⁴) / (2.27×10⁻⁴) = (0.30/0.10)ʸ ⇒ 3 = 3ʸ ⇒ y = 1.Comparing Exp 3 and Exp 1 (keeping [B] constant):(13.62×10⁻⁴) / (6.81×10⁻⁴) = (0.20/0.10)ˣ ⇒ 2 = 2ˣ ⇒ x = 1.So, the rate law is r₀ = k[A][B].Using Exp 2 to find k: 2.27×10⁻⁴ = k[0.10][0.10] = k(0.01) ⇒ k = 2.27×10⁻² L mol⁻¹ s⁻¹.Now, calculate r₀ for [A]=0.50 mol/L and [B]=0.50 mol/L:r₀ = (2.27×10⁻²) × [0.50] × [0.50] = (2.27×10⁻²) × 0.25 = 0.5675×10⁻² = 5.675×10⁻³ mol L⁻¹ s⁻¹.Converting to 10⁻⁴ notation: 56.75 × 10⁻⁴ mol L⁻¹ s⁻¹.

Answer

56.75 x 10-4 mol L-1s-1

Answer

5.675 x 10-4 mol L-1s-1

Answer

21.67 x 10-4 mol L-1s-1

Answer

2.167 x 10-4 mol L-1s-1

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