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, let's go through each part step by step:
1.) What is the activation energy of the reverse reaction?
To find the activation energy of the reverse reaction, you need to remember that the activation energy remains the same regardless of the direction of the reaction. Therefore, the activation energy for the reverse reaction will also be 35.0 kJ/mol, which is the same as the forward reaction.
b.) What is the value of enthalpy of reaction for the reverse reaction?
To determine the enthalpy of reaction for the reverse reaction, you can use the fact that the enthalpy change for the reverse reaction is the negative of the enthalpy change for the forward reaction. In this case, the enthalpy of the reverse reaction would be -20 kJ, which means the reaction is exothermic.
c.) What is the effect of an increase in temperature on the rate of reaction and the value of the rate constant?
An increase in temperature generally leads to an increase in the rate of reaction. This is explained by the kinetic molecular theory, which states that at higher temperatures, molecules have more kinetic energy and move faster, resulting in more frequent and energetic collisions, increasing the rate of reaction. The rate constant, however, may follow a different trend. In general, the rate constant increases with increasing temperature, following the Arrhenius equation, which you mentioned: k = Ae^(-E_a/RT). Here, R is the gas constant and T is the temperature in Kelvin. As the temperature increases, the exponential term e^(-E_a/RT) becomes larger, causing the rate constant to increase.
d.) To calculate the value of the rate constant for the reaction at a higher temperature, you can use the Arrhenius equation mentioned above. Plug in the given values: R = 8.314 J/(mol*K), T1 = 25°C = 298 K, T2 = 35°C = 308 K, and the activation energy E_a = 35.0 kJ/mol. Then, you can calculate the ratio k2/k1 to find the rate constant at the higher temperature.
e.) The addition of a homogeneous catalyst affects both the rate of reaction and the value of the rate constant. A catalyst increases the rate of reaction by providing an alternate reaction pathway with a lower activation energy. This allows more reactant molecules to overcome the activation barrier and participate in the reaction, thereby increasing the reaction rate. However, a catalyst does not change the value of the rate constant, as the rate constant depends only on temperature and the nature of the reacting species, not on the presence of a catalyst.
I hope this helps clarify your questions! Let me know if you need any further assistance or any specific mathematical calculations.