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Temperature and activation energy

1.[3p]

A reaction has k=3.38×10-5 s⁻¹ at 298.15 K and 4.87×10-3 s⁻¹ at 338.15 K. What is Ea, in kJ mol⁻¹?

CorrectNot quite: 104

2.[2p]

On an Arrhenius plot of lnk against 1/T, what is the intercept?

Correct
The answer is: $\ln A$
The answer is: $\ln A$
The answer is: $\ln A$

3.[3p]

A catalyst lowers the activation energy by 27 kJ mol⁻¹. By what factor does the rate rise at 298 K, taking A as unchanged and RT=2.479 kJ mol⁻¹?

CorrectNot quite: 54000

4.[1p]

A catalyst increases the equilibrium yield of a reaction as well as its rate.

The answer is: False
Correct

5.[3p]

Why can a catalyst not shift the position of equilibrium?

Correct
The answer is: It lowers the forward and reverse barriers equally, so the ratio of rate constants is unchanged
The answer is: It lowers the forward and reverse barriers equally, so the ratio of rate constants is unchanged
The answer is: It lowers the forward and reverse barriers equally, so the ratio of rate constants is unchanged

6.[2p]

Collision theory predicts a pre-exponential factor of 7.4×1011 dm³ mol⁻¹ s⁻¹ for a reaction whose measured value is 1.24×106. What does the discrepancy show?

Correct
The answer is: Molecules must meet in a particular orientation, which featureless spheres cannot represent
The answer is: Molecules must meet in a particular orientation, which featureless spheres cannot represent
The answer is: Molecules must meet in a particular orientation, which featureless spheres cannot represent

7.[2p]

For an elementary step, Ea forward is 130 kJ mol⁻¹ and Ea reverse is 175 kJ mol⁻¹. What is ΔH for the step, in kJ mol⁻¹?

CorrectNot quite: -45

8.[3p]

Which statements about the transition state are correct?

Select all that apply

Correct
Correct
Correct
The answer is: It sits at a maximum of energy along the reaction coordinate, Old bonds are partly broken and new ones partly formed in it, It is not an intermediate and cannot be isolated

9.[2p]

Match each quantity to what a plot against 1/T measures.

  • Slope of ln k against 1/T

  • Slope of ln K against 1/T

  • Intercept of ln k against 1/T

  • minus the activation energy over R

  • the log of the prefactor

  • minus the reaction enthalpy over R

Show the answer

Slope of ln k against 1/T: minus the activation energy over R Slope of ln K against 1/T: minus the reaction enthalpy over R Intercept of ln k against 1/T: the log of the prefactor