Using three equations and their enthalpies

Problems 11 - 20

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

N _{2}(g) + 4H_{2}(g) + Cl_{2}(g) ---> 2NH_{4}Cl(s)ΔH _{rxn}= -630.78 kJN _{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)ΔH _{rxn}= -296.4 kJ

Determine the identity of the two missing products and calculate the ΔH_{rxn} for this reaction:

NH_{4}Cl(s) ---> _____ + _____

**Solution:**

1) Write out all three data equations:

^{1}⁄_{2}H_{2}(g) +^{1}⁄_{2}Cl_{2}(g) ---> HCl(g)ΔH _{rxn}= -92.3 kJ/molN _{2}(g) + 4H_{2}(g) + Cl_{2}(g) ---> 2NH_{4}Cl(s)ΔH _{rxn}= -630.78 kJN _{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)ΔH _{rxn}= -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 NH_{4}Cl 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}⁄_{2}H_{2}(g) +^{1}⁄_{2}Cl_{2}(g) ---> HCl(g)ΔH _{rxn}= -92.3 kJ/mol2NH _{4}Cl(s) ---> N_{2}(g) + 4H_{2}(g) + Cl_{2}(g)ΔH _{rxn}= +630.78 kJN _{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)ΔH _{rxn}= -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?

N_{2}and H_{2}= do not work because no Cl

N_{2}and Cl_{2}= no H

N_{2}, H_{2}, Cl_{2}= that's three products

Here is what works:

NH_{3}and HCl

4) What that means is we (1) divide the third data equation by two, in order to get one NH_{3} as a product and (2) divide the second data equation by 2, in order to get one NH_{4}Cl as a reactant. Here it is:

^{1}⁄_{2}H_{2}(g) +^{1}⁄_{2}Cl_{2}(g) ---> HCl(g)ΔH _{rxn}= -92.3 kJ/molNH _{4}Cl(s) --->^{1}⁄_{2}N_{2}(g) + 2H_{2}(g) +^{1}⁄_{2}Cl_{2}(g)ΔH _{rxn}= +315.39 kJ^{1}⁄_{2}N_{2}(g) +^{3}⁄_{2}H_{2}(g) ---> NH_{3}(g)ΔH _{rxn}= -148.2kJ

5) When you add the three equations, the 2H_{2} , ^{1}⁄_{2}Cl_{2} and ^{1}⁄_{2}N_{2} all cancel, leaving only this:

NH_{4}Cl ---> NH_{3}+ 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) + Fgiven:_{2}(g) ---> ClF_{3}(g)

2CIF(g) + O _{2}(g) ---> Cl_{2}O(g) + F_{2}O(g)ΔH = 167.4 kJ 2CIF _{3}(g) + 2O_{2}(g) ---> Cl_{2}O(g) + 3F_{2}O(g)ΔH = 341.4 kJ 2F _{2}(g) + O_{2}(g) ----> 2F_{2}O(g)ΔH = -43.4 kJ

**Solution:**

1) Adding equations 1 and 3:

2ClF + 2F _{2}+ 2O_{2}---> Cl_{2}O + 3F_{2}OΔH = 167.4 + (-43.4) = 124 kJ

2) Reverse equation 2:

Cl _{2}O + 3F_{2}O ---> 2O_{2}+ 2ClF_{3}ΔH = - (341.4) = -341.4 kJ

3) Now add the above two new equations:

Common terms (Cl_{2}O, 3F_{2}O and 2O_{2}) on either side of the reaction will cancelled.

2ClF + 2F _{2}---> 2ClF_{3}ΔH = 124 + (-341.4) = -217.4

4) Divide by 2 to get the final equation:

ClF(g) + F _{2}(g) ---> ClF_{3}(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(NO_{3})_{2}(s) from the following data.

8Mg(s) + Mg(NO _{3})_{2}(s) ---> Mg_{3}N_{2}(s) + 6MgO(s)ΔH° = -3280.88 Mg _{3}N_{2}(s) ---> 3Mg(s) + N_{2}(g)ΔH° = +461.08 2MgO(s) ---> 2Mg(s) + O _{2}(g)ΔH° = +1203.60

**Solution**

1) The chemical reaction for the ΔH°_{f} for Mg(NO_{3})_{2}(s) is this:

Mg(s) + N_{2}(g) + 3O_{2}(g) ---> Mg(NO_{3})_{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(NO_{3})_{2} is on the right-hand side. Like this:

Mg_{3}N_{2}(s) + 6MgO(s) ---> 8Mg(s) + Mg(NO_{3})_{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) + N _{2}(g) ---> Mg_{3}N_{2}(s)ΔH° -461.08 2Mg(s) + O _{2}(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) + 3O_{2}(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:

Mg _{3}N_{2}(s) + 6MgO(s) ---> 8Mg(s) + Mg(NO_{3})_{2}(s)ΔH° = +3280.88 3Mg(s) + N _{2}(g) ---> Mg_{3}N_{2}(s)ΔH° = -461.08 6Mg(s) + 3O _{2}(g) ---> 6MgO(s)ΔH° = -3610.80

6) When you add them up, the Mg_{3}N_{2}, 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:

2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(ℓ)ΔH = -571.6 kJ N _{2}O_{5}(g) + H_{2}O(ℓ) ---> 2HNO_{3}(ℓ)ΔH = -73.7 kJ ^{1}⁄_{2}N_{2}(g) +^{3}⁄_{2}O_{2}(g) +^{1}⁄_{2}H_{2}(g) ---> HNO_{3}(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:

N_{2}+^{5}⁄_{2}O_{2}---> N_{2}O_{5}

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:

H _{2}O(ℓ) ---> H_{2}(g) +^{1}⁄_{2}O_{2}(g)ΔH = +285.8 kJ 2HNO _{3}---> N_{2}O_{5}(g) + H_{2}O(ℓ)ΔH = +73.7 kJ N _{2}(g) + 3O_{2}(g) + H_{2}(g) ---> 2HNO_{3}(ℓ)ΔH = -348.2 kJ

4) What cancels?

H_{2}O - equation 1 and 2

H_{2}- equation 1 and 3

2HNO_{3}- equation 2 and 3^{1}⁄_{2}O_{2}- equation 1 and 3This last cancel will reduce the O

_{2}from^{6}⁄_{2}to^{5}⁄_{2}, which is what we want.

5) Add up the enthalpies:

(+285.8) + (+73.7) + (-348.2) = +11.3 kJThis source gives +11.3 kJ for the enthalpy of formation for N

_{2}O_{5}(g).

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

2C(graphite) + 3 H_{2}(g) ---> C_{2}H_{6}(g)

from the data given below:

C(graphite) + O _{2}(g) ---> CO_{2}(g)ΔH = -393.5 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH = -285.8 kJ 2C _{2}H_{6}(g) + 7O_{2}(g) ---> 4CO_{2}(g) + 6H_{2}O(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 3H_{2}, so multiply second data equation by 3.

c) The third data equation is divided by 2 and reversed. This is to put one C_{2}H_{6}on the product side.

2) The reason we do not reverse the first two equations is because the C and the H_{2} are already on the reactant side, where we want them. if we do everything correctly, all the CO_{2}, O_{2} and H_{2}O will cancel out.

2C(graphite) + 2O _{2}(g) ---> 2CO_{2}(g)ΔH = -787.0 kJ 3H _{2}(g) +^{3}⁄_{2}O_{2}(g) ---> 3H_{2}O(ℓ)ΔH = -857.4 kJ 2CO _{2}(g) + 3H_{2}O(ℓ) ----> C_{2}H_{6}(g) +^{7}⁄_{2}O_{2}(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:** Construct Hess's Law cycle to calculate the standard enthalpy change for the reaction:

2C(s, gr) + 2H_{2}(g) + O_{2}(g) ---> CH_{3}COOH

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

C(s, gr) = -394; H_{2}(g) = -286; CH_{3}COOH = -876

**Solution:**

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

C(s, gr) + O_{2}---> CO_{2}; ΔH = -394 kJ

H_{2}+^{1}⁄_{2}O_{2}---> H_{2}O; ΔH = -286 kJ

CH_{3}COOH + 2O_{2}---> 2CO_{2}+ 2H_{2}O; ΔH = -876 kJSince 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) + 2O_{2}---> 2CO_{2}; ΔH = -788 kJ <--- multiplied by two, enthalpy also mult. by 2

2H_{2}+ O_{2}---> 2H_{2}O; ΔH = -572 kJ <--- multiplied by two, enthalpy also mult. by 2

2CO_{2}+ 2H_{2}O ---> CH_{3}COOH + 2O2; ΔH = +876 kJ <--- reversed the equation, enthalpy changes sign

3) When we add the three equations, as modified above, we find the 2CO_{2} cancels out, as well as 2H_{2}O 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, C_{2}H_{5}OH(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 CO_{2}(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) + 3H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> C_{2}H_{5}OH(ℓ); ΔH = -278 kJNote 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) + O_{2}(g) ---> CO_{2}(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) + 3H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> C_{2}H_{5}OH(ℓ); ΔH = -278 kJ

C_{2}H_{5}OH(ℓ) + 3O_{2}(g) ---> 2CO_{2}(s) + 3H_{2}O(ℓ); ΔH = -1368 kJ

H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(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) + 3H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> C_{2}H_{5}OH(ℓ); ΔH = -278 kJ

C_{2}H_{5}OH(ℓ) + 3O_{2}(g) ---> 2CO_{2}(s) + 3H_{2}O(ℓ); ΔH = -1368 kJ

3H_{2}O(ℓ) ---> 3H_{2}(g) +^{3}⁄_{2}O_{2}(g); ΔH = +858 kJ <--- reversed equation, multiplied by 3The first two data equations are left untouched.

5) Adding the three equations and their enthalpies gives:

2C(s, gr) + 2O_{2}(g) ---> 2CO_{2}(g); ΔH = -788 kJ

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

C(s,gr) + O_{2}(g) ---> CO_{2}(g); ΔH = -394 kJThis link gives the value for the standard enthalpy of formation for carbon dioxide.

**Problem #18:** Calculate the enthalpy of reaction for:

CH_{4}(g) + 4Cl_{2}(g) ---> CCl4(g) + 4HCl(g)

Use the following reactions and given ΔH′s.

C(s) + 2H_{2}(g) ---> CH_{4}(g) ΔH = -74.6 kJ

C(s) + 2Cl_{2}(g) ---> CCl_{4}(g) ΔH = -95.7 kJ

H_{2}(g) + Cl_{2}(g) ---> 2HCl(g) ΔH = -184.6 kJ

**Solution**

1) Manipulate the three data equations as follows:

reverse equation 1: CH_{4}---> C + 2H_{2}ΔH = +74.6 kJ

keep equation 2 as is: C + 2Cl_{2}---> CCl_{4}ΔH = -95.7 kJ

multiply equation 3 by 2: H_{2}+ 2Cl_{2}---> 4HCl ΔH = -369.2 kJ

2) Add the three equations:

C + CH_{4}+ 2Cl_{2}+ 2H_{2}+ 2Cl_{2}---> C + 2H_{2}+ CCl_{4}+ 4HCl

3) Cancel what appears on both sides to obtain:

CH_{4}+ 4Cl_{2}---> CCl_{4}+ 4HClwhich is the target equation.

4) Add the enthalpies for the answer:

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

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

C_{12}H_{22}O_{11}(s) + 12O_{2}(g) ---> 12CO_{2}(g) + 11H_{2}O(ℓ) ΔH = -1349.6 kcalH

_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ) ΔH = -68.3 kcalC(s) + O

_{2}(g) ---> CO_{2}(g) ΔH = -94 kcal

Determine the heat of formation for sucrose.

**Solution:**

1) The heat of formation is this reaction:

12C(s) + 11H_{2}(g) +^{11}⁄_{2}O_{2}(g) ---> C_{12}H_{22}O_{11}(s)

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

12CO_{2}(g) + 11H_{2}O(ℓ) ---> C_{12}H_{22}O_{11}(s) + 12O_{2}(g) ΔH = +1349.6 kcal <---reversed the equation11H

_{2}(g) +^{11}⁄_{2}O_{2}(g) ---> 11H_{2}O(ℓ) ΔH = -751.3 kcal <--- multiplied by 1112C(s) + 12O

_{2}(g) ---> 12CO_{2}(g) ΔH = -1128 kcal <--- multiplied by 12

3) When the three data equations are added, the 12CO_{2} will cancel out as will the 11H_{2}O and the 12O_{2}. The ^{11}⁄_{2}O_{2} will remain. Add the three ΔH values for the final answer.

**Problem #20:** Given:

CH_{4}+ 2O_{2}---> CO_{2}+ 2H_{2}O ΔH = -820 kJ/ molCH

_{4}+ CO_{2}---> 2CO + 2H_{2}ΔH = 206 kJ/molCH

_{4}+ H_{2}O ---> CO + 3H_{2}ΔH = 247 kJ/mol

Calculate the enthalpy of reaction for:

2CH_{4}+ 3O_{2}---> 2CO + 4H_{2}O

**Solution:**

1) Apply the following changes to the data equations:

a) multiply by 3

b) multiply by 3

c) reverse equation, multiply by 2The need for a 3 and 2 is because the hyrogens in equations 2 & 3 have coefficients of 2 and 3. Since we need the H

_{2}to cancel, we need to set up the equations to have six (the LCM of 2 and 3) H_{2}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 CO

_{2}.

2) The result:

3CH_{4}+ 6O_{2}-> 3CO_{2}+ 6H_{2}O ΔH = -2460 kJ/ mol3CH

_{4}+ 3CO_{2}---> 6CO + 6H_{2}ΔH = 618 kJ/mol2CO + 6H

_{2}---> 2CH_{4}+ 2H_{2}O ΔH = -494 kJ/mol

3) Add and cancel:

4CH_{4}+ 6O_{2}---> 4CO + 8H_{2}O ΔH = -2336 kJ

4) Divide by 2:

2CH_{4}+ 3O_{2}---> 2CO + 4H_{2}O ΔH = -1168 kJ