Reaction 1: NaOH + HCL --> H20 + NaCl delta H of -100.332 kj/mol
Reaction 2: NaOH + NH4Cl --> NH3 + H20 : delta H of 358.639 kj/mol
Reaction 3: HCl + NH3 --> NH4Cl : delta H of -51.701 kj/mol
Use your answers from question 2 above and Hess's law to determine the experimental molar enthalpy for Reaction 3.
I've already found the theoretical enthalpy changes of each reaction, but I'm confused about how to find the experimental.
Other information: Reaction 3 had temp change of 5.8 degrees C and absorbed 2500 joules of heat energy.
Thank you so much!
3 answers
It appears to me that you haven't posted all of the information or at least I don't know what's going on. Is this an experiment you are conducting? I don't understand what you have determined; ie., the "theoretical enthalpy changes of each reaction." Are the delta H values listed for each the values you have determined experimentally?
Sorry if I was being unclear. Yes, this is an experiment that I conducted. We calculated the temperature change between one reactant and when the reaction occurred.
The first reaction had a temp change of 12.5, the 2nd had 1.10, and the third had 5.8 degrees C. I've also calculated q (amount of heat energy) for each of the reactions written above. Furthermore, I calculated the theoretical enthalpy change for each of these reactions.
The question that I am having trouble with is how to calculate the experimental molar enthalpy for reaction 3 using the theoretical enthalpy change and Hess's law. Thanks again!
The first reaction had a temp change of 12.5, the 2nd had 1.10, and the third had 5.8 degrees C. I've also calculated q (amount of heat energy) for each of the reactions written above. Furthermore, I calculated the theoretical enthalpy change for each of these reactions.
The question that I am having trouble with is how to calculate the experimental molar enthalpy for reaction 3 using the theoretical enthalpy change and Hess's law. Thanks again!
OK. What you want is delta H for the equation of NH3 + HCl ==> NH4Cl
You get that by taking the equation for reaction 1 and adding the reverse of the equation for reaction 2 as shown below. But before I do that I want to point out that equation 2 is not correct. Perhaps you just didn't type all of the products. You see you are missing Na^+ and Cl^- on the product side. What it should be is
NaOH + NH4Cl ==> NaCl + NH3 + H2O
NaOH + HCl ==> NaCl + H2O
NH3 + H2O + NaCl ==> NH4Cl + NaOH
----------------------------
add to get
NaOH in #1 and #2 cancel(opposite sides)
H2O in #1 and #2 cancel(opposite sides)
NaCl is #1 & #2 cancel(opposite sides)
You are left with
NH3 + HCl ==> NH4Cl which is what you wanted.
For the delta H data, you take dH for #1, change the sign on #2, add them and you will have delta H for the equation shown in 3. Sorry it took me so long but I had that pesky NaCl in the reaction and tried to find how to make it cancel. It took awhile for me to recognize that the initial equation you had was short the NaCl.
You get that by taking the equation for reaction 1 and adding the reverse of the equation for reaction 2 as shown below. But before I do that I want to point out that equation 2 is not correct. Perhaps you just didn't type all of the products. You see you are missing Na^+ and Cl^- on the product side. What it should be is
NaOH + NH4Cl ==> NaCl + NH3 + H2O
NaOH + HCl ==> NaCl + H2O
NH3 + H2O + NaCl ==> NH4Cl + NaOH
----------------------------
add to get
NaOH in #1 and #2 cancel(opposite sides)
H2O in #1 and #2 cancel(opposite sides)
NaCl is #1 & #2 cancel(opposite sides)
You are left with
NH3 + HCl ==> NH4Cl which is what you wanted.
For the delta H data, you take dH for #1, change the sign on #2, add them and you will have delta H for the equation shown in 3. Sorry it took me so long but I had that pesky NaCl in the reaction and tried to find how to make it cancel. It took awhile for me to recognize that the initial equation you had was short the NaCl.
2 answers