In a reaction 2HI → H2 + I2, the concentration of HI decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. What is the rate of reaction during this interval?

Question

In a reaction 2HI → H2 + I2, the concentration of HI decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. What is the rate of reaction during this interval? (A) 5 × 10-3 M min-1 (B) 2.5 × 10-3 M min- 1 (C) 5 × 10-2 M min-1 (D) 2.5 × 10-2 M min-1

Tardigrade Admin · Accepted Answer

Average rate =-(1/2) (Δ[R]/Δ t)=-(1/2)×(0.4-0.5/10) =(1/2)×(0.1/10)=5×10-3 M min-1

Q. In a reaction $2 H I \to H_{2} + I_{2}$ , the concentration of $H I$ decreases from $0.5 m o l L^{- 1}$ to $0.4 m o l L^{- 1}$ in $10$ minutes. What is the rate of reaction during this interval?

$5 \times 1 0^{- 3} M mi n^{- 1}$

$2.5 \times 1 0^{- 3} M mi n^{- 1}$

$5 \times 1 0^{- 2} M mi n^{- 1}$

$2.5 \times 1 0^{- 2} M mi n^{- 1}$

Average rate $= - \frac{1}{2} \frac{Δ [ R ]}{Δ t} = - \frac{1}{2} \times \frac{0.4 - 0.5}{10}$
$= \frac{1}{2} \times \frac{0.1}{10} = 5 \times 1 0^{- 3} M mi n^{- 1}$

Q. In a reaction 2HI→H2​+I2​, the concentration of HI decreases from 0.5molL−1 to 0.4molL−1 in 10 minutes. What is the rate of reaction during this interval?

5×10−3Mmin−1

2.5×10−3Mmin−1

5×10−2Mmin−1

2.5×10−2Mmin−1