A reaction rate depends on the concentration of reactant B according to the rate law: Rate = k[B]^(3/2). If the concentration of B is doubled, by what factor does the rate increase?

✅ Correct. Your selected answer is correct. The initial rate law is Rate₁ = k[B]^(3/2). When the concentration of B is doubled, the new rate is Rate₂ = k(2[B])^(3/2) = k · 2^(3/2) · [B]^(3/2). The ratio of the new rate to the initial rate is Rate₂Rate₁ = 2^(3/2). Calculating the value: 2^(3/2) = 2√(2) ≈ 2 × 1.414 = 2.828. This value is approximately 2.8, which matches your selection.

A reaction rate depends on the concentration of reactant B according to the rate law: Rate = k[B]^(3/2). If the concentration of B is doubled, by what factor does the rate increase?

✅ Correct. Your selected answer is correct. The initial rate law is Rate₁ = k[B]^(3/2). When the concentration of B is doubled, the new rate is Rate₂ = k(2[B])^(3/2) = k · 2^(3/2) · [B]^(3/2). The ratio of the new rate to the initial rate is Rate₂Rate₁ = 2^(3/2). Calculating the value: 2^(3/2) = 2√(2) ≈ 2 × 1.414 = 2.828. This value is approximately 2.8, which matches your selection.

Solved: A reaction rate depends on the concentration of…

✅ Correct.

Your selected answer is correct.

The initial rate law is $\text{Rate}_1 = k[B]^{3/2}$ .
When the concentration of B is doubled, the new rate is $\text{Rate}_2 = k(2[B])^{3/2} = k \cdot 2^{3/2} \cdot [B]^{3/2}$ .
The ratio of the new rate to the initial rate is $\frac{\text{Rate}_2}{\text{Rate}_1} = 2^{3/2}$ .
Calculating the value: $2^{3/2} = 2\sqrt{2} \approx 2 \times 1.414 = 2.828$ .
This value is approximately $\boxed{2.8}$ , which matches your selection.

A reaction rate depends on the concentration of reactant B according to the rate law: Rate = k[B]^(3/2). If the concentration of B is doubled, by what factor does the rate increase?

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