Question

Use the equation of the line shown in the graph below to calculate the activation energy, Ea in kJ/mol, and the frequency factor A (units are s?). Report your answers 3 significant figures. Upload your answers and work. No work = no credit. The equation of the line y = – 1340.1x + 27.644 corresponds to: ln k = (-Ea / R) (1 / T) + ln A y = -1340.1x + 27.644 ln k vs 1/T 1/T (K?)

          Use the equation of the line shown in the graph below to calculate the activation energy, Ea in kJ/mol, and the frequency factor A (units are s?). Report your answers 3 significant figures. Upload your answers and work. No work = no credit.

The equation of the line y = – 1340.1x + 27.644 corresponds to:

ln k = (-Ea / R) (1 / T) + ln A

y = -1340.1x + 27.644
ln k vs 1/T
1/T (K?)

Show more…

Use the equation of the line shown in the graph below to calculate the activation energy, Ea in kJ/mol, and the frequency factor A (units are s?). Report your answers 3 significant figures. Upload your answers and work. No work = no credit.

The equation of the line y = – 1340.1x + 27.644 corresponds to:

ln k = (-Ea / R) (1 / T) + ln A

y = -1340.1x + 27.644
ln k vs 1/T
1/T (K?)

Added by Inmaculada L.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

03/14/2023

Step 1

The Arrhenius equation is given by: $k = Ae^{\frac{-Ea}{RT}}$ Taking the natural logarithm of both sides, we get: $\ln k = \ln A - \frac{Ea}{R} \cdot \frac{1}{T}$ Comparing this equation with the given line equation, we can see that: $y = -\frac{Ea}{R}x + \ln Show more…

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Use the equation of the line shown in the graph below to calculate the activation energy, Ea in kJ/mol, and the frequency factor A (units are s^-1). Report your answers 3 significant figures. Upload your answers and work. No work = no credit. The equation of the line y = -1340.1x + 27.644 corresponds to: ln k = (-Ea/R)(1/T) + ln A

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