### Hess' Law of Constant Heat SummationUsing three equations and their enthalpiesProblems 11 - 20

Problem #11: The standard enthalpy of formation for hydrogen chloride is −92.3 kJ/mol. Given it and the following data:

 N2(g) + 4H2(g) + Cl2(g) ---> 2NH4Cl(s) ΔHrxn = −630.78 kJ N2(g) + 3H2(g) ---> 2NH3(g) ΔHrxn = −296.4 kJ

Determine the identity of the two missing products and calculate the ΔHrxn for this reaction:

NH4Cl(s) ---> _____ + _____

Solution:

1) Write out all three data equations:

 1⁄2H2(g) + 1⁄2Cl2(g) ---> HCl(g) ΔHrxn = −92.3 kJ/mol N2(g) + 4H2(g) + Cl2(g) ---> 2NH4Cl(s) ΔHrxn = −630.78 kJ N2(g) + 3H2(g) ---> 2NH3(g) ΔHrxn = −296.4kJ

2) I know that the second data equation will have to be flipped and divided by 2. This is because I know only NH4Cl is on the reactant side in the target equation. Let's do that (the flip only, the division by 2 happens below) and keep everything else the same:

 1⁄2H2(g) + 1⁄2Cl2(g) ---> HCl(g) ΔHrxn = −92.3 kJ/mol 2NH4Cl(s) ---> N2(g) + 4H2(g) + Cl2(g) ΔHrxn = +630.78 kJ N2(g) + 3H2(g) ---> 2NH3(g) ΔHrxn = −296.4kJ

3) I know that I have two (and only two) missing products. What two substances in the data equations will make two products and involve N, H, and Cl?

N2 and H2 = do not work because no Cl
N2 and Cl2 = no H
N2, H2, Cl2 = that's three products

Here is what works:

NH3 and HCl

4) What that means is we (1) divide the third data equation by two, in order to get one NH3 as a product and (2) divide the second data equation by 2, in order to get one NH4Cl as a reactant. Here it is:

 1⁄2H2(g) + 1⁄2Cl2(g) ---> HCl(g) ΔHrxn = −92.3 kJ/mol NH4Cl(s) ---> 1⁄2N2(g) + 2H2(g) + 1⁄2Cl2(g) ΔHrxn = +315.39 kJ 1⁄2N2(g) + 3⁄2H2(g) ---> NH3(g) ΔHrxn = −148.2kJ

5) When you add the three equations, the 2H2 , 12Cl2 and 12N2 all cancel, leaving only this:

NH4Cl ---> NH3 + HCl

Add the three enthalpies for the final answer of −82.805 kJ. Rounded off, −82.8 kJ

Problem #12: Calculate ΔH for this reaction:

ClF(g) + F2(g) ---> ClF3(g)
given:
 2CIF(g) + O2(g) ---> Cl2O(g) + F2O(g) ΔH = 167.4 kJ 2CIF3(g) + 2O2(g) ---> Cl2O(g) + 3F2O(g) ΔH = 341.4 kJ 2F2(g) + O2(g) ----> 2F2O(g) ΔH = −43.4 kJ

Solution:

1) Adding equations 1 and 3:

 2ClF + 2F2 + 2O2 ---> Cl2O + 3F2O ΔH = 167.4 + (-43.4) = 124 kJ

2) Reverse equation 2:

 Cl2O + 3F2O ---> 2O2 + 2ClF3 ΔH = −341.4 kJ

3) Now add the above two new equations:

Common terms (Cl2O, 3F2O and 2O2) on either side of the reaction will cancelled.

 2ClF + 2F2 ---> 2ClF3 ΔH = 124 + (−341.4) = −217.4

4) Divide by 2 to get the final equation:

 ClF(g) + F2(g) ---> ClF3(g) ΔH = −108.7

NOTE: For every operation that you perform on the equation, perform the same on ΔH values too!

Comment: I copied this from Yahoo Answers since it is not the normal ChemTeam style of presentation. I thought you might be interested in how another brain approaches a solution to these types of problems.

Problem #13: Calculate the ΔH°f for Mg(NO3)2(s) from the following data.

 8Mg(s) + Mg(NO3)2(s) ---> Mg3N2(s) + 6MgO(s) ΔH° = −3280.88 Mg3N2(s) ---> 3Mg(s) + N2(g) ΔH° = +461.08 2MgO(s) ---> 2Mg(s) + O2(g) ΔH° = +1203.60

Solution

1) The chemical reaction for the ΔH°f for Mg(NO3)2(s) is this:

Mg(s) + N2(g) + 3O2(g) ---> Mg(NO3)2(s)

2) We need to rearrange the three data equations to yield the above reaction when added together. Look at the first equation and see that it must be reversed so that the Mg(NO3)2 is on the right-hand side. Like this:

Mg3N2(s) + 6MgO(s) ---> 8Mg(s) + Mg(NO3)2(s) ΔH° = +3280.88

3) Notice the 8Mg. We have to make that go away while leaving one Mg on the left- hand side. Look at the second and third data equations. We will have to flip both and somehow create 9Mg on the left-hand side. Here's the flip:

 3Mg(s) + N2(g) ---> Mg3N2(s) ΔH° −461.08 2Mg(s) + O2(g) ---> 2MgO(s) ΔH° −1203.60

4) I'm going to multiply the third data equation (the second equation in the list just above) by 3:

6Mg(s) + 3O2(g) ---> 6MgO(s) ΔH° −3610.80

5) That gives me 9Mg from the 3Mg in the second data equation and 6Mg in the third. Here are all three revised data equations:

 Mg3N2(s) + 6MgO(s) ---> 8Mg(s) + Mg(NO3)2(s) ΔH° = +3280.88 3Mg(s) + N2(g) ---> Mg3N2(s) ΔH° = −461.08 6Mg(s) + 3O2(g) ---> 6MgO(s) ΔH° = −3610.80

6) When you add them up, the Mg3N2, the 6MgO and 8Mg will cancel. Add up the three enthalpies for the final answer:

+3280.88 + (−461.08) + (−3610.80) = −791 kJ

The −791 value is mentioned here.

Problem #14: Given the following thermochemical equations:

 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH = −571.6 kJ N2O5(g) + H2O(ℓ) ---> 2HNO3(ℓ) ΔH = −73.7 kJ 1⁄2N2(g) + 3⁄2O2(g) + 1⁄2H2(g) ---> HNO3(ℓ) ΔH = −174.1 kJ

Calculate ΔH for the formation of one mole of dinitrogen pentoxide from its elements in their standard state at 25 °C and 1 atm.

Solution:

1) Here's the target equation:

N2 + 52O2 ---> N2O5

2) Here's what you need to do:

1) divide equation by 2 and flip
2) flip second eq
3) multiply equation by 2

3) Here's the result:

 H2O(ℓ) ---> H2(g) + 1⁄2O2(g) ΔH = +285.8 kJ 2HNO3 ---> N2O5(g) + H2O(ℓ) ΔH = +73.7 kJ N2(g) + 3O2(g) + H2(g) ---> 2HNO3(ℓ) ΔH = −348.2 kJ

4) What cancels?

H2O ---> equation 1 and 2
H2 ---> equation 1 and 3
2HNO3 ---> equation 2 and 3
12O2 ---> equation 1 and 3

This last cancel will reduce the O2 from 62 to 52, which is what we want.

(+285.8) + (+73.7) + (−348.2) = +11.3 kJ

This source gives +11.3 kJ for the enthalpy of formation for N2O5(g).

Problem #15: Calculate the enthalpy change for the reaction:

2C(s, gr) + 3H2(g) ---> C2H6(g)

from the data given below:

 C(s, gr) + O2(g) ---> CO2(g) ΔH = −393.5 kJ H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = −285.8 kJ 2C2H6(g) + 7O2(g) ---> 4CO2(g) + 6H2O(ℓ) ΔH = −3119.6 kJ

Solution:

1) We need to rearrange the three data equations so that, when they are added together, the target equation emerges.

a) We know we need 2C, so the first data equation will be multiplied by 2.
b) We know we need 3H2, so multiply second data equation by 3.
c) The third data equation is divided by 2 and reversed. This is to put one C2H6 on the product side.

2) The reason we do not reverse the first two equations is because the C and the H2 are already on the reactant side, where we want them. if we do everything correctly, all the CO2, O2 and H2O will cancel out.

 2C(s, gr) + 2O2(g) ---> 2CO2(g) ΔH = −787.0 kJ 3H2(g) + 3⁄2O2(g) ---> 3H2O(ℓ) ΔH = −857.4 kJ 2CO2(g) + 3H2O(ℓ) ----> C2H6(g) + 7⁄2O2(g) ΔH = 1559.8 kJ

Note what happened to the enthalpies: multiplied or divided as well as the sign change when I reversed the third data equation.

3) To obtain the final answer, add up the three enthalpy values from the changed data equations.

This problem from Yahoo Answers asks for the ΔH to be calculated for 32 °C rather than the usual standard temperature value of 25 °C. The answerer first calculates the ΔH at 25 °C and then converts it to the value it would be at 32 °C.

Problem #16: Use Hess' Law to calculate the standard enthalpy change for the reaction:

2C(s, gr) + 2H2(g) + O2(g) ---> CH3COOH

use the following standard enthalpies of combustion at 298 K (given in kJ mol-1)

C(s, gr) = −394; H2(g) = −286; CH3COOH = −876

Solution:

1) First, we write out all the combustion reactions:

C(s, gr) + O2 ---> CO2; ΔH = −394 kJ
H2 + 12O2 ---> H2O; ΔH = −286 kJ
CH3COOH + 2O2 ---> 2CO2 + 2H2O; ΔH = −876 kJ

Since the enthalpies were given in kJ mol-1, I made sure to balance each equation with only one mole of the first substance in each equation.

2) Next, we modify the equations in order to reach the desired target equation:

2C(s, gr) + 2O2 ---> 2CO2; ΔH = −788 kJ <--- multiplied by two, enthalpy also mult. by 2
2H2 + O2 ---> 2H2O; ΔH = −572 kJ <--- multiplied by two, enthalpy also mult. by 2
2CO2 + 2H2O ---> CH3COOH + 2O2; ΔH = +876 kJ <--- reversed the equation, enthalpy changes sign

3) When we add the three equations, as modified above, we find the 2CO2 cancels out, as well as 2H2O and two of the three oxygens. After canceling, we recover our target equation.

4) The enthalpy of the target equation is this:

−788 + (−572) + 876 = −484 kJ

This link goes to the NIST Chemistry Webbook listing for acetic acid, where the value for its standard enthalpy of formation is given.

Problem #17: (a) The standard enthalpy of formation of ethanol, C2H5OH(ℓ), is −278 kJ mol¯1. Write a thermochemical equation which represents the standard enthalpy of formation of ethanol.

(b) Use the above information in part (a) and the following data to calculate the standard enthalpy of formation of CO2(g).

Standard enthalpy of combustion of ethanol = −1368 kJ mol¯1
Standard enthalpy of combustion of hydrogen = −286 kJ mol¯1

Solution

1) Solution to part (a):

2C(s, gr) + 3H2(g) + 12O2(g) ---> C2H5OH(ℓ); ΔH = −278 kJ

Note that the standard state of carbon is graphite, not any of its other allotropes (such as diamond or buckminsterfullerene).

2) The equation for the formation of carbon dioxide is this:

C(s, gr) + O2(g) ---> CO2(g)

We need to determine the enthalpy for it.

3) The first step in the solution to part (b) is to write all three data equations:

 2C(s, gr) + 3H2(g) + 1⁄2O2(g) ---> C2H5OH(ℓ) ΔH = −278 kJ C2H5OH(ℓ) + 3O2(g) ---> 2CO2(s) + 3H2O(ℓ) ΔH = −1368 kJ H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = −286 kJ

4) We have to modify the data equations so as to recover the target equation when we add them together:

 2C(s, gr) + 3H2(g) + 1⁄2O2(g) ---> C2H5OH(ℓ) ΔH = −278 kJ C2H5OH(ℓ) + 3O2(g) ---> 2CO2(s) + 3H2O(ℓ) ΔH = −1368 kJ 3H2O(ℓ) ---> 3H2(g) + 3⁄2O2(g) ΔH = +858 kJ <--- reversed equation, multiplied by 3

The first two data equations are left untouched.

5) Adding the three equations and their enthalpies gives:

2C(s, gr) + 2O2(g) ---> 2CO2(g); ΔH = −788 kJ

6) Divide by two to get a formation equation (which only shows one mole of the product):

C(s,gr) + O2(g) ---> CO2(g); ΔH = −394 kJ

This link gives the value for the standard enthalpy of formation for carbon dioxide.

Problem #18: Calculate the enthalpy of reaction for:

CH4(g) + 4Cl2(g) ---> CCl4(g) + 4HCl(g)

Use the following reactions and given ΔH′s.

 C(s) + 2H2(g) ---> CH4(g) ΔH = −74.6 kJ C(s) + 2Cl2(g) ---> CCl4(g) ΔH = −95.7 kJ H2(g) + Cl2(g) ---> 2HCl(g) ΔH = −184.6 kJ

Solution

1) Manipulate the three data equations as follows:

 CH4 ---> C + 2H2 ΔH = +74.6 kJ <--- reversed the equation C + 2Cl2 ---> CCl4 ΔH = −95.7 kJ <--- no change 2H2 + 2Cl2 ---> 4HCl ΔH = −369.2 kJ <--- multiply equation by 2

C + CH4 + 2Cl2 + 2H2 + 2Cl2 ---> C + 2H2 + CCl4 + 4HCl

3) Cancel what appears on both sides to obtain:

CH4 + 4Cl2 ---> CCl4 + 4HCl

which is the target equation.

+74.6 + (−95.7) + (−369.2) = −390.3 kJ

Problem #19: Some heats of combustion are given below:

 C12H22O11(s) + 12O2(g) ---> 12CO2(g) + 11H2O(ℓ) ΔH = −1349.6 kcal H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = −68.3 kcal C(s, gr) + O2(g) ---> CO2(g) ΔH = −94 kcal

Determine the heat of formation for sucrose.

Solution:

1) The heat of formation for sucrose is this reaction:

12C(s, gr) + 11H2(g) + 112O2(g) ---> C12H22O11(s)

2) Let us manipulate the three data equations as follows:

 12CO2(g) + 11H2O(ℓ) ---> C12H22O11(s) + 12O2(g) ΔH = +1349.6 kcal <---reversed the equation 11H2(g) + 11⁄2O2(g) ---> 11H2O(ℓ) ΔH = −751.3 kcal <--- multiplied by 11 12C(s, gr) + 12O2(g) ---> 12CO2(g) ΔH = −1128 kcal <--- multiplied by 12

3) When the three data equations are added, the 12CO2 will cancel out as will the 11H2O and the 12O2. The 112O2 will remain. Add the three ΔH values for the final answer.

Problem #20: Given:

 CH4 + 2O2 ---> CO2 + 2H2O ΔH = −820 kJ CH4 + CO2 ---> 2CO + 2H2 ΔH = 206 kJ CH4 + H2O ---> CO + 3H2 ΔH = 247 kJ

Calculate the enthalpy of reaction for:

2CH4 + 3O2 ---> 2CO + 4H2O

Solution:

1) Apply the following changes to the data equations:

a) multiply by 3
b) multiply by 3
c) reverse equation, multiply by 2

The need for a 3 and 2 is because the hyrogens in equations 2 & 3 have coefficients of 2 and 3. Since we need the H2 to cancel, we need to set up the equations to have six (the LCM of 2 and 3) H2 on each side.

Multiplying eq 2 by 3 means we must also multiply eq 1 by 3. We have to do this in order to cancel the CO2.

2) The result:

 3CH4 + 6O2 ---> 3CO2 + 6H2O ΔH = −2460 kJ 3CH4 + 3CO2 ---> 6CO + 6H2 ΔH = 618 kJ 2CO + 6H2 ---> 2CH4 + 2H2O ΔH = −494 kJ

4CH4 + 6O2 ---> 4CO + 8H2O ΔH = −2336 kJ

4) Divide by 2:

2CH4 + 3O2 ---> 2CO + 4H2O ΔH = −1168 kJ

Problem #21: Calculate the enthalpy of combustion of methane, given that the standard enthalpies of formation of methane, carbon dioxide, water are −74.80, −393.50 and −285.83 (all in kJ/mol)?

Solution:

1) First, we will write the equation be asked for, the combustion of methane:

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(ℓ)

2) Next, we write the three formation equations and their enthalpies:

 C(s, gr) + 2H2(g) ---> CH4(g) ΔH = −74.80 kJ C(s, gr) + O2(g) ---> CO2(g) ΔH = −393.50 kJ H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = −285.83 kJ

3) Now manipulate the three data equations in order to create the combustion equation for methane:

 CH4(g) ---> C(s, gr) + 2H2(g) ΔH = +74.80 kJ <--- flip this equation, note sign change on enthalpy C(s, gr) + O2(g) ---> CO2(g) ΔH = −393.50 kJ <--- leave untouched 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH = −571.66 kJ <--- multiply by 2

4) When the three equations are added, the C(s, gr) and the 2H2(g) will cancel, leaving:

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(ℓ)     ΔH = −890.36 kJ