I don't understand how to approach this question in the textbook. Can someone please explain it to me?
Thank you!
Ch. 8 Question 8.61 (Hess's Law)
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 Joined: Fri Jul 22, 2016 3:00 am
Re: Ch. 8 Question 8.61 (Hess's Law)
Hello,
So this problem is dealing with Hess's Law, that is manipulating equations and their delta H's to give the desired chemical equation. There are multiple ways to think about this but I will lead you through mine. I recommend you to write it out on paper and follow each step because it is pretty convoluted at some times to follow.
1.) I see that we need 1 H2 in the final equation. I also notice that equation 2 and equation 3 both have H2 in their reactions and their coefficients differ by 1. So, in order to make it so that 43 = 1, I reverse equation 2. This will leave 1 H2 on the reactant side, which is what we need.
2.) By reversing equation 2, I have also made it such that there are N2 in the products of equation 2 and N2 in the reactants of equation 3. Since there isn't any N2 in the final equation, I can go ahead and cancel these out.
3.) Next, I decide to cancel out the NH4Br since it is not in the final equation. To do this I notice that there is 2 NH4Br in the products of equation 3 and 1 NH4Br in the products of equation 1. To cancel these out, I reverse equation 1 and multiply by 2 such that there are now 2 NH4Br in both the product and reactant sides.
4.) By doing the above manipulations, I notice that I can cancel out the 2NH3 on the product side of equation 1 and the reactant side of 2. Since it is not in the final equation, I carry this cancellation out accordingly.
5.) Summing up all the remaining reactants and products from the 3 equations will give us the desired equation. With the desired equation, all we need to do now is to take the manipulations we did with the equations and apply them to the delta H's to calculate the final delta H for the reaction we want.
Summary: flip sign and multiply by 2 for delta H of equation 1, giving 376.64 kJ, flip sign for delta H of equation 2, giving 92.22 kJ. Nothing was done to equation 3 so that remains the same as given, i.e 541.66 kJ. Take the summation of these 3 values yield 72.80 kJ as the enthalpy change for the desired reaction, which is also the answer the solution gives.
So this problem is dealing with Hess's Law, that is manipulating equations and their delta H's to give the desired chemical equation. There are multiple ways to think about this but I will lead you through mine. I recommend you to write it out on paper and follow each step because it is pretty convoluted at some times to follow.
1.) I see that we need 1 H2 in the final equation. I also notice that equation 2 and equation 3 both have H2 in their reactions and their coefficients differ by 1. So, in order to make it so that 43 = 1, I reverse equation 2. This will leave 1 H2 on the reactant side, which is what we need.
2.) By reversing equation 2, I have also made it such that there are N2 in the products of equation 2 and N2 in the reactants of equation 3. Since there isn't any N2 in the final equation, I can go ahead and cancel these out.
3.) Next, I decide to cancel out the NH4Br since it is not in the final equation. To do this I notice that there is 2 NH4Br in the products of equation 3 and 1 NH4Br in the products of equation 1. To cancel these out, I reverse equation 1 and multiply by 2 such that there are now 2 NH4Br in both the product and reactant sides.
4.) By doing the above manipulations, I notice that I can cancel out the 2NH3 on the product side of equation 1 and the reactant side of 2. Since it is not in the final equation, I carry this cancellation out accordingly.
5.) Summing up all the remaining reactants and products from the 3 equations will give us the desired equation. With the desired equation, all we need to do now is to take the manipulations we did with the equations and apply them to the delta H's to calculate the final delta H for the reaction we want.
Summary: flip sign and multiply by 2 for delta H of equation 1, giving 376.64 kJ, flip sign for delta H of equation 2, giving 92.22 kJ. Nothing was done to equation 3 so that remains the same as given, i.e 541.66 kJ. Take the summation of these 3 values yield 72.80 kJ as the enthalpy change for the desired reaction, which is also the answer the solution gives.

 Posts: 14
 Joined: Wed Sep 21, 2016 2:56 pm

 Posts: 13
 Joined: Wed Sep 21, 2016 2:55 pm
Re: Ch. 8 Question 8.61 (Hess's Law)
We would use this approach assuming we were not given something like the Appendix data regarding enthalpies of formation right? Or are these two different concepts.

 Posts: 10
 Joined: Fri Jul 22, 2016 3:00 am
Re: Ch. 8 Question 8.61 (Hess's Law)
Hello Riley,
Yes. We would use this approach if they ask you to use Hess's Law to find the enthalpy change associated with a reaction they want. In order to do this, the enthalpy change of the other reactions they want you to manipulate should be given (like how it was given in this problem).
The enthalpies of formation technique is just another way of calculating the enthalpy change of a reaction. Just like how there are many ways to get to Westwood from campus, by foot or by car or by skateboard, there are multiple ways to get the enthalpy change of a reaction depending on what the problem gives you. Usually the problem will specify what method they like as we have seen in some of the textbook problems or it is clear from what they give you. Like if the reaction enthalpies were given but not the appendix with the enthalpies of formation, than it pretty clear that in order to solve the problem, we would have to use Hess's Law with those reaction enthalpy values.
Yes. We would use this approach if they ask you to use Hess's Law to find the enthalpy change associated with a reaction they want. In order to do this, the enthalpy change of the other reactions they want you to manipulate should be given (like how it was given in this problem).
The enthalpies of formation technique is just another way of calculating the enthalpy change of a reaction. Just like how there are many ways to get to Westwood from campus, by foot or by car or by skateboard, there are multiple ways to get the enthalpy change of a reaction depending on what the problem gives you. Usually the problem will specify what method they like as we have seen in some of the textbook problems or it is clear from what they give you. Like if the reaction enthalpies were given but not the appendix with the enthalpies of formation, than it pretty clear that in order to solve the problem, we would have to use Hess's Law with those reaction enthalpy values.
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