An endothermic reaction was found to have an activation energy of 100 kJ. If this reaction is reversed, the activation energy is only 20 kJ. What is the value of delta H for the endothermic reaction?
For an endothermic reaction, Ea forward = Ea reverse + delta E rxn.
So the answer would be positive 120 kJ?
No. If Ea forward is 100 and Ea reverse is 20, then delta E reaction is 80.
I looked for and count not find a good diagram. Finally I succeeded.
http://www.kentchemistry.com/links/Kinetics/PEDiagrams.htm
You can see that 40 + 30 = total = 70 for exothermic and equally you see that 100 = total = 20 + 80
http://www.kentchemistry.com/links/Kinetics/PEDiagrams.htm
To determine the value of ΔH for an endothermic reaction, we need to use the relationship between activation energy and ΔH, as described by the Arrhenius equation:
k = Ae^(-Ea/RT)
Where:
- k is the rate constant of the reaction
- A is the pre-exponential factor
- Ea is the activation energy
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
When the reaction is reversed, the sign of ΔH changes, so we can compare the activation energies given to calculate the change in enthalpy.
Let's assume that the temperatures and the pre-exponential factor are constant for both the forward and reverse reactions. We can write the equation for the forward reaction as:
k(forward) = A * e^(-Ea(forward)/RT)
And the equation for the reverse reaction as:
k(reverse) = A * e^(-Ea(reverse)/RT)
Since the activation energy for the forward reaction is 100 kJ and for the reverse reaction is 20 kJ, we can rewrite the equations as:
k(forward) = A * e^(-100,000/RT)
k(reverse) = A * e^(-20,000/RT)
Now, we can find the ratio of the rate constants:
k(forward)/k(reverse) = [A * e^(-100,000/RT)] / [A * e^(-20,000/RT)]
The pre-exponential factor cancels out, and we are left with:
k(forward)/k(reverse) = e^(-100,000/RT + 20,000/RT)
To find the change in enthalpy, we can take the natural logarithm of both sides of the equation:
ln(k(forward)/k(reverse)) = -80,000/(RT)
The left-hand side of the equation gives us the natural logarithm of the rate constant ratio. The right-hand side simplifies to -80,000/(RT).
Thus, the change in enthalpy, ΔH, is equal to -80,000 J/mol (or -80 kJ/mol) for the given endothermic reaction.