Question

The instantanous rate of Dissaperance MnO4- ion in the following reaction is 4.56 x 10^-3 ms

2 MnO^-4 + 10 I^- + 16H^+ gives 2Mn^+2 + 5 I2 + 8H2O

The rate of apperance of I2 is

(a) 1.14 x 10^-3
(b) 5.7×10^-3
(c) 4.56x10^-4
(d) 1. 14x10^-2​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Rate \: of \: appearance \: of \: I_{2} =1.14 \times {10}^{ - 2} \: M {s}^{ - 1}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies 2 \: MnO_{4}^{ - } + 10 {I}^{ - } + 16H^{ + } \to 2 \: Mn^{2 + } + 5 \: I_{2} + 8 \: H_{2}O \\ \\ \tt: \implies Rate \: of \: dissapearance \: of \:[ MnO_{4} ^{ - }] = 4.56 \times {10}^{ - 3} \: M {s}^{ - 1} \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies Rate \: of \: appearance \: of \: [I_{2} ]= ?$

• According to given question :

$\bold{As \: we \: know \:that} \\ \tt: \implies \frac{[MnO_{4}^{ - }] }{2} = \frac{[ I_{2}] }{5} \\ \\ \tt: \implies \frac{4.56 \times {10}^{ - 3} }{2} = \frac{[I_{2}]}{5} \\ \\\tt: \implies [I_{2} ]= \frac{4.56 \times {10}^{ - 3} \times 5}{2} \\ \\ \tt: \implies [ I_{2} ]= \frac{22.8 \times 10^{ - 3} }{2} \\ \\ \tt: \implies [I_{2} ]=11.4 \times {10}^{ - 3} \\ \\ \green{\tt: \implies [I_{2}] =1.14 \times {10}^{ - 2} \: M {s}^{ -1 } } \\ \\ \green{\tt \therefore Rate \: of \: appearance \: of \: [ I_{2}] \: is \: 1.14 \times {10}^{ - 2} \: M {s}^{ - 1} }$

Answer 2

GIVEN:

• 2MnO-₄ + 10 I- + 16H+ + 2Mn+₂ + 5I₂ + H₂O
• Rate of disappearance of MnO-₄ : 4.56 ×10-³

TO FIND:

• Rate of disappearance of I₂ (Iodide)

SOLUTION:

We know that

(MnO-₄)/2 = I₂/5

→ 4.56 × 10-³/2 = I₂/5

→ 4.56 × 10-³ × 5/2 = I₂

→ 22.8 × 10-³/2

→ 11.4 × 10-³

It can be written as

→ 1.14 × 10-²

Hence , rate of appearance of I₂ = 1.14 × 10-²