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Question: In a reaction \(2HI \rightarrow H_{2} + I_{2},\) the concentration of HI decrease from \(0.5molL^{- ...

In a reaction 2HI→H2+I2,2HI \rightarrow H_{2} + I_{2}, the concentration of HI decrease from 0.5molL−10.5molL^{- 1} to 0.4molL−10.4molL^{- 1} in 10 minutes what is the rate of reaction during this interval?

A

5×10−3Mmin⁡−15 \times 10^{- 3}M\min^{- 1}{}

B

2.5×10−3Mmin⁡−12.5 \times 10^{- 3}M\min^{- 1}{}

C

5×10−2Mmin⁡−15 \times 10^{- 2}M\min^{- 1}{}

D

2.5×10−2Mmin⁡−12.5 \times 10^{- 2}M\min^{- 1}{}

Answer

5×10−3Mmin⁡−15 \times 10^{- 3}M\min^{- 1}{}

Explanation

Solution

Average rate =−12Δ[R]Δt=−12×0.4−0.510= - \frac{1}{2}\frac{\Delta\lbrack R\rbrack}{\Delta t} = - \frac{1}{2} \times \frac{0.4 - 0.5}{10}

=12×0.110=5×10−3Mmin⁡−1= \frac{1}{2} \times \frac{0.1}{10} = 5 \times 10^{- 3}M\min^{- 1}{}