I'm just overly confused on how to begin this especially with the fractions... please show step by step if you could. Quiz very Soon!
Example A:
H2(g)+1/2 O2 (g) ----> H2O(l) ... ΔH= -285.8 kJ
N2O5(g) + H2O(l) -----> 2HNO3(l) ΔH= -76.56 kJ
1/2 N2(g) + 3/2 O2(g) +1/2 H2(g) -----> ΔH= -174.1 kJ
Calculate ΔH for the reaction... 2N2(g) + 5O2(g) -----> 2N2O5(g)
Example B:
Fe2O3(s)+3CO(g) -----> 2Fe(s) + 3CO2(g) ...ΔH=-28 kJ
3 Fe2O3(s) + CO(g) -----> 2Fe3O4(s) + CO2 (g) ...ΔH=-59 kJ
Fe3O4(s) +CO(g) -----> 3FeO(s) +CO2 (g) ΔH=+38 kJ
Calculate ΔH: FeO(s) + CO (g) ----->Fe(s) +CO2(g)
To calculate the enthalpy change (ΔH) for a given reaction, you need to use the Hess's Law. Hess's Law states that the ΔH for an overall reaction is the sum of the ΔH of the individual reactions involved.
Let's break down both examples step by step:
Example A:
1. First, we need to look at the given reactions and identify any common species. In this case, we have the reaction:
1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) --> ?
This reaction has N2, O2, and H2 as common species to the desired reaction: 2N2(g) + 5O2(g) --> 2N2O5(g)
2. To match the desired reaction, we need to manipulate the given reactions. We can multiply the first reaction by 4 to balance the number of nitrogen atoms:
2 N2(g) + 6 O2(g) + 2 H2(g) --> ?
3. We can also multiply the second reaction by 2 to balance the number of nitrogen and oxygen atoms:
2 N2O5(g) + 2 H2O(l) --> 4 HNO3(l)
4. Add the manipulated reactions together:
2 N2(g) + 6 O2(g) + 2 H2(g) + 2 N2O5(g) + 2 H2O(l) --> 4 HNO3(l)
5. The given reactions have ΔH values associated with them. Add up the ΔH values for the individual reactions to get the overall ΔH:
ΔH = -174.1 kJ + (-285.8 kJ) + (-76.56 kJ)
ΔH = -536.46 kJ
Therefore, the ΔH for the reaction 2 N2(g) + 5 O2(g) --> 2 N2O5(g) is -536.46 kJ.
Example B:
1. Analyze the given reactions and identify any common species. Here, the reaction needed is:
FeO(s) + CO(g) --> Fe(s) + CO2(g)
The common species are FeO and CO.
2. Manipulate the given reactions to match the desired reaction. We will multiply the first reaction by 3, and the second reaction by 2:
3 Fe2O3(s) + 9 CO(g) --> 6 Fe(s) + 9 CO2(g)
6 Fe2O3(s) + 2 CO(g) --> 4 Fe3O4(s) + 6 CO2(g)
3. Add the manipulated reactions together:
6 Fe2O3(s) + 9 CO(g) + 6 Fe2O3(s) + 2 CO(g) --> 6 Fe(s) + 9 CO2(g) + 4 Fe3O4(s) + 6 CO2(g)
12 Fe2O3(s) + 11 CO(g) --> 6 Fe(s) + 10 Fe3O4(s) + 15 CO2(g)
4. The given reactions have ΔH values associated with them. Add up the ΔH values for the individual reactions:
ΔH = (-28 kJ) + (-59 kJ) + (+38 kJ)
ΔH = -49 kJ
Therefore, the ΔH for the reaction FeO(s) + CO(g) --> Fe(s) + CO2(g) is -49 kJ.
Remember to always manipulate the given reactions to match the desired reaction, and then sum up the ΔH values to find the overall ΔH for the reaction.
To calculate ΔH for the given reactions, you need to use the concept of Hess's Law. Hess's Law states that the total ΔH for a chemical reaction is equal to the sum of the ΔH values for the individual reactions that make up the overall reaction, as long as you manipulate the reactions to match the desired overall reaction.
Let's go through the steps for each example:
Example A:
1. Given reactions:
- Reaction 1: H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.8 kJ
- Reaction 2: N2O5(g) + H2O(l) → 2HNO3(l) ΔH = -76.56 kJ
- Reaction 3: 1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) → ΔH = -174.1 kJ
2. We need to manipulate the given reactions to obtain the desired overall reaction: 2N2(g) + 5O2(g) → 2N2O5(g)
3. Multiply Reaction 1 by 2 to obtain 2 moles of H2O:
2[H2(g) + 1/2 O2(g) → H2O(l)] ΔH = -571.6 kJ
4. Multiply Reaction 2 by 2 to obtain 2 moles of HNO3:
2[N2O5(g) + H2O(l) → 2HNO3(l)] ΔH = -153.12 kJ
5. Flip Reaction 3 and multiply it by 2 to obtain 2 moles of N2 and 5 moles of O2:
2[1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) →] ΔH = 348.2 kJ
6. Add the manipulated reactions together:
2[H2(g) + 1/2 O2(g) → H2O(l)] + 2[N2O5(g) + H2O(l) → 2HNO3(l)] + 2[1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) →]
ΔH = -571.6 kJ - 153.12 kJ + 348.2 kJ
ΔH = -376.52 kJ
Therefore, ΔH for the reaction 2N2(g) + 5O2(g) → 2N2O5(g) is -376.52 kJ.
Example B:
1. Given reactions:
- Reaction 1: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -28 kJ
- Reaction 2: 3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g) ΔH = -59 kJ
- Reaction 3: Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ΔH = +38 kJ
2. We need to manipulate the given reactions to obtain the desired overall reaction: FeO(s) + CO(g) → Fe(s) + CO2(g)
3. Multiply Reaction 1 by 3 to obtain 3 moles of CO2:
3[Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)] ΔH = -84 kJ
4. Multiply Reaction 2 by 2 to obtain 2 moles of Fe3O4:
2[3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g)] ΔH = -118 kJ
5. Flip Reaction 3 to obtain 1 mole of FeO:
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ΔH = -38 kJ
6. Add the manipulated reactions together:
3[Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)] + 2[3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g)] + [Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g)]
ΔH = -84 kJ - 118 kJ - 38 kJ
ΔH = -240 kJ
Therefore, ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g) is -240 kJ.