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Question: In a reaction \(2X \rightarrow Y\) the concentration of X decreases from \(0.50M\) to \(0.38M\) in 1...

In a reaction 2X→Y2X \rightarrow Y the concentration of X decreases from 0.50M0.50M to 0.38M0.38M in 10 min. what is the rate of reaction in Ms−1Ms^{- 1}during this interval?

A

2×10−42 \times 10^{- 4}

B

4×10−24 \times 10^{- 2}

C

2×10−22 \times 10^{- 2}

D

1×10−21 \times 10^{- 2}

Answer

2×10−42 \times 10^{- 4}

Explanation

Solution

Rate of reaction =Δ[X]Δt= \frac{\Delta\lbrack X\rbrack}{\Delta t}

Δ[X]=Xi−Xf=0.50−0.38=0.12M\Delta\lbrack X\rbrack = X_{i} - X_{f} = 0.50 - 0.38 = 0.12M

Rate =0.1210×60=2×10−4Ms−1= \frac{0.12}{10 \times 60} = 2 \times 10^{- 4}Ms^{- 1}