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Question: A substance \('A'\) decomposes by a first order reaction initially with \((a) = 2.00mol\) and after ...

A substance ′A′'A' decomposes by a first order reaction initially with (a)=2.00mol(a) = 2.00mol and after

200min, 6mu(a−x)=0.15mol200{min,}\mspace{6mu}(a - x) = 0.15mol. For this reaction what is the value of kk

A

1.29×10−2min⁡−11.29 \times 10^{- 2}\min^{- 1}{}

B

2.29×10−2min⁡−12.29 \times 10^{- 2}\min^{- 1}{}

C

3.29×10−2min⁡−13.29 \times 10^{- 2}\min^{- 1}{}

D

4.40×10−2min⁡−14.40 \times 10^{- 2}\min^{- 1}{}

Answer

1.29×10−2min⁡−11.29 \times 10^{- 2}\min^{- 1}{}

Explanation

Solution

Given [a]=2.00mol,\lbrack a\rbrack = 2.00mol, t=200minutet = 200\text{minute} and

(a−x)=0.15mol(a - x) = 0.15mol

k=2.303tlog⁡10a(a−x)=2.303200log⁡102.000.15\mathbf{k =}\frac{\mathbf{2.303}}{\mathbf{t}}\mathbf{\log}_{\mathbf{10}}\frac{\mathbf{a}}{\mathbf{(a - x)}}\mathbf{=}\frac{\mathbf{2.303}}{\mathbf{200}}\mathbf{\log}_{\mathbf{10}}\frac{\mathbf{2.00}}{\mathbf{0.15}} =1.29×10−2min⁡−1= 1.29 \times 10^{- 2}\min^{- 1}{}