The temperature coefficient of a reaction is 2.5. If its rate constant at T1K is 2.5×10−3s−1, the rate constant at T2K in s−1 is T2>T1

The temperature coefficient is the number of times the rate constant increases when the temperature is increased by 10oC.It is given to us that T2 is higher than T1. Therefore, we can write that KT2KT1=2.5Where, the rate constant at temperature T2 kelvin is denoted by KT2 and the rate constant at temperatureT1 kelvin is denoted by KT1.KT22.5×10-3s-1=2.5 KT2=6.25×10-3 s-1

The temperature coefficient of a reaction is 2.5. If its rate constant at T1K is 2.5×10−3s−1, the rate constant at T2K in s−1 is T2>T1

The temperature coefficient is the number of times the rate constant increases when the temperature is increased by 10oC.It is given to us that T2 is higher than T1. Therefore, we can write that KT2KT1=2.5Where, the rate constant at temperature T2 kelvin is denoted by KT2 and the rate constant at temperatureT1 kelvin is denoted by KT1.KT22.5×10-3s-1=2.5 KT2=6.25×10-3 s-1

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The temperature coefficient of a reaction is 2.5. If its rate constant at $T_{1} K is 2.5 \times 10^{- 3} s^{- 1}$ , the rate constant at $T_{2} K in s^{- 1} is (T_{2} > T_{1})$

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1.0×10^–3

6.25×10^–3

1.0×10^–2

6.25×10^–2

answer is B.

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Detailed Solution

The temperature coefficient is the number of times the rate constant increases when the temperature is increased by 10^oC.
It is given to us that T₂ is higher than T₁.
Therefore, we can write that $\frac{K_{T_{2}}}{K_{T_{1}}} = 2.5$
Where, the rate constant at temperature $T_{2}$ kelvin is denoted by $K_{T_{2}}$ and the rate constant at temperature $T_{1}$ kelvin is denoted by $K_{T_{1}}$ .
$\frac{K_{T_{2}}}{2.5 \times 10^{- 3} s^{- 1}} = 2.5 K_{T_{2}} = 6.25 \times 10^{- 3} s^{- 1}$