Doing review for an exam and I can't find how to do this problem in my notes.
2 SO2 + 2 Cl2 -----> 2 SOCl2 + O2
from a previous problem I solved:
rate=k[SO2]^2[Cl2]
k = 467.666666... or in sig. figs 467
Energy reaction:
T= 25 C
The activation energy (Ea) is 35.0kJ/mole
enthalpy of reaction (^Hrxn) is +20 kJ.
Questions:
1.) What is the activation energy of the reverse reaction?
Is this just -35 kJ/mole? If not, how do I determine?
b.) What is the value of enthalpy of reaction for the reverse reaction?
Can't find equation.
c.) What is the effect an increase in temperature would have on the rate of the reaction and on the value of the rate constant? Explain the manner in which it affects the rate and the rate constant.
Ah, a qualitative question. I believe that an increase in temp. would cause an increase in the rate of reaction b/c an increase in heat speeds up molecules. Not sure about the rate constant.
d.) Calculate the value of the rate constant for this same reaction if the temperature increased from 25 C to 35 C.
Not sure of the equation here.
e.) What is the effect the addition of a homogeneous catalyst has on the rate of a reaction and the value of a rate constant? Explain the manner in which it affects the rate and the rate constant.
I'm guessing since this is a catalyst it speeds up the reaction (by definition of a catalyst, right?). Once again, not sure about the rate constant.
Okay, I'm guessing from what I've read there is a single equation I don't have that I need.
I do have an equation kind of like what is going on here:
k=Ae^(-Ea/RT) with:
R= 8.314 J/K mol (or 8.314*10^(-3) kJ/K mol)
I'm just not sure how enthalpy of reaction is involved here, so I think I need something else.
Thank you.
To answer your questions, you can use the concepts of the Arrhenius equation and the principle of microscopic reversibility.
1.) To find the activation energy of the reverse reaction, you can use the principle of microscopic reversibility. Since the forward reaction has an activation energy (Ea) of 35.0 kJ/mol, the reverse reaction will have an activation energy of the same magnitude but opposite in sign. Therefore, the activation energy of the reverse reaction is -35.0 kJ/mol.
2.) The enthalpy of reaction for the reverse reaction can be determined using the principle of microscopic reversibility. Since the enthalpy of the forward reaction (^Hrxn) is +20 kJ, the enthalpy of the reverse reaction will have the same magnitude but opposite in sign. Therefore, the value of the enthalpy of reaction for the reverse reaction is -20 kJ.
3.) An increase in temperature generally increases the rate of a reaction. This is because an increase in temperature leads to an increase in the kinetic energy of molecules, causing more frequent and energetic collisions, which in turn increases the rate of successful collisions and therefore the rate of the reaction. The rate constant (k) is also affected by temperature. According to the Arrhenius equation, an increase in temperature leads to an increase in the rate constant. This is because the exponential term in the Arrhenius equation, e^(-Ea/RT), becomes larger with a higher temperature, resulting in a larger value for the rate constant.
4.) To calculate the value of the rate constant for the same reaction at a different temperature, you can use the Arrhenius equation. The equation you have, k = A * e^(-Ea/RT), is the correct equation. Here, A is the pre-exponential factor, R is the gas constant, T is the temperature in Kelvin, and Ea is the activation energy. Plug in the given values, including the new temperature of 35°C (which should be converted to Kelvin by adding 273), and calculate the new value of the rate constant (k).
5.) The addition of a homogeneous catalyst increases the rate of a reaction by providing an alternative pathway with a lower activation energy. By lowering the activation energy, the catalyst allows more reactant molecules to overcome the energy barrier and participate in the reaction, thereby increasing the rate. However, adding a catalyst does not affect the value of the rate constant (k). The rate constant is determined by the nature of the reactants and the temperature, and it remains unchanged in the presence of a catalyst.
I hope this helps! Let me know if you have any further questions.