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 10^{- 3} M m i n^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.5 \times 10^{- 3} M m i n^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 \times 10^{- 2} M m i n^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5 \times 10^{- 2} M m i n^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $5 \times 10^{- 3} M m i n^{- 1}$
Average rate $= - \frac{1}{2} \frac{△ [R]}{△ t} = - \frac{1}{2} \times \frac{0.4 - 0.5}{10}$

$= \frac{1}{2} \times \frac{0.1}{10} = 5 \times 10^{- 3} M m i n^{- 1}$ .

Suggest Corrections

Similar questions

In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval?

Q. Define complex reaction. In a reaction,

2 A \to P r o d u c t s

, the concentration of A decreases from

0.5 m o l L^{- 1}

0.4 m o l L^{- 1}

in 10 minutes. Calculate the rate during this interval.

Q. In a reaction;

2 A \to P r o d u c t s

, the concentration of

A

decreases from

0.5 m o l l i t r e^{- 1}

0.4 m o l l i t r e^{- 1}

10

minutes. Calculate rate during this interval.

Q. 1) For the reaction

R ⟶ P

,the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
2) In a reaction,

2 A ⟶ P r o d u c t s

, the concentration of

A

decreases from

0.5

0.4 m o l L^{- 1}

in 10 minutes. Calculate the rate during this interval.

In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 min. The rate during this interval will be

Join BYJU'S Learning Program

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?

The correct option is A 5×10−3M min−1Average rate =−12△[R]△t=−12×0.4−0.510=12×0.110=5×10−3M min−1.

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?

The correct option is A $5 \times 10^{- 3} M m i n^{- 1}$
Average rate $= - \frac{1}{2} \frac{△ [R]}{△ t} = - \frac{1}{2} \times \frac{0.4 - 0.5}{10}$

$= \frac{1}{2} \times \frac{0.1}{10} = 5 \times 10^{- 3} M m i n^{- 1}$ .