If the rate of reaction is 2.6 × 1 0 − 3 mol L − 1 s − 1 at 5 0 0 C and 7.02 × 1 0 − 2 mol L − 1 s − 1 at 8 0 0 C ; then what will be the temperature coefficient of the reaction?

r 1 r 2 = ( μ ) Δ T /10 2.6 × 1 0 − 3 7.02 × 1 0 − 2 = ( μ ) 30/10 27 = μ 3 μ = 3

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ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minQB

ChemistryMediuminteger

If the rate of reaction is $2.6 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$ at $5 0^{0} C$ and $7.02 \times 1 0^{- 2} mol L^{- 1} s^{- 1}$ at $8 0^{0} C$ ; then what will be the temperature coefficient of the reaction?

Answer:

Solution:

$\frac{r _{2}}{r _{1}} = (μ)^{Δ T /10}$

$\frac{7.02 \times 1 0 ^{- 2}}{2.6 \times 1 0 ^{- 3}} = (μ)^{30/10}$

$27 = μ^{3}$

$μ = 3$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

⏱ 2mℹ️ Source: QB

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