For the reaction, Cl2 + 2I- longrightarrow I2 + 2Cl- , the initial concentration of I- was 0.20 mol L-1 and the concentration after 20 min was 0.18 mol L-1. Then the rate of formation of I2 in mol L-1 min-1 would be

Question

For the reaction, Cl2 + 2I- longrightarrow I2 + 2Cl- , the initial concentration of I- was 0.20 mol L-1 and the concentration after 20 min was 0.18 mol L-1. Then the rate of formation of I2 in mol L-1 min-1 would be (A) 1 × 10-3 (B) 5 × 10-4 (C) 1 × 10-4 (D) 2 × 10-3

Tardigrade Admin · Accepted Answer

Rate of formation of I2 = (dI2/dt) = -(1/2) (d [I-]/dt) = (1/2) × (0.20 - 0.18/20) = (1/2) × (0.02/20) = 5 × 10-4 mol L-1 min-1

Q. For the reaction, $C l_{2} + 2 I^{-} ⟶ I_{2} + 2 C l^{-},$ the initial concentration of $I^{-}$ was 0.20 mol $L^{- 1}$ and the concentration after 20 min was 0.18 mol $L^{- 1}$ . Then the rate of formation of $I_{2}$ in mol $L^{- 1} mi n^{- 1}$ would be

$1 \times 1 0^{- 3}$

$5 \times 1 0^{- 4}$

$1 \times 1 0^{- 4}$

$2 \times 1 0^{- 3}$

$5 \times 1 0^{- 3}$

Rate of formation of $I_{2} = \frac{d I _{2}}{d t} = - \frac{1}{2} \frac{d [ I ^{-} ]}{d t}$
$= \frac{1}{2} \times \frac{0.20 - 0.18}{20} = \frac{1}{2} \times \frac{0.02}{20}$
$= 5 \times 1 0^{- 4} m o l L^{- 1} mi n^{- 1}$

Q. For the reaction, Cl2​+2I−⟶I2​+2Cl−, the initial concentration of I− was 0.20 mol L−1 and the concentration after 20 min was 0.18 mol L−1. Then the rate of formation of I2​ in mol L−1min−1 would be

1×10−3

5×10−4

1×10−4

2×10−3

5×10−3

Rate of formation of I2​=dtdI2​​=−21​dtd[I−]​ =21​×200.20−0.18​=21​×200.02​ =5×10−4molL−1min−1