Question

Calculate delta Hrxn for the following reaction:
5C(s)+6H2(g)--->C5H12(l)

use the following reactions and given delta H's:

C5H12(l)+8O2(g)---> 5CO2(g)+6H2O(g) delta H= -3505.8 kJ
C(s)+O2(g)--->CO2(g) delta H= -393.5 kJ
2H2(g)+O2(g)---> 2H2O(g) delta H= -483.5 kJ

The steps to solving this problem would be greatly appreciated! Thanks so much!

Answer 1

5C(s)+6H2(g)C5H12(1) C5H12(1)+802(g)-> 5CO2(g)+6H2O(g) delta H--3505.8 kJ C(s)+O2(g)CO2(g) delta H-393.5 kJ 2H2(g)+02(g2H20(g) delta H- -483.5 kJ Hess law: The heat of reaction is independent of the number of steps in which the reaction takes place. That is, the net heat of reaction is equal to the sum of the heats of reaction of each step The C5H12(1) is the product and we need 1 mol. So, the first step of the reaction is, 5CO2(g)+6H2O(g) C5H12(1)+802(g) delta H+3505.8 kJ We need 5 moles of C. So, the second step is, 5x(C(s)+O2(g)->CO2(g)) delta H 5x-393.5 kJ 5C(s)+502(g)5CO2(g)) delta H -1967.5 KJ We need 6 moles of H2. So, the final step is, 3x(2H2(g)+O2(g) 2H20(g)delta H- 3x-483.5 kJ 6H2(g)+302(6H20(g)) delta H--1450.5 kJ

If we add these steps, we will get the global reaction, 5CQ2(g)+6N2O(g) > C5H12(1)+802(g) delta H +3505.8 kJ 5C(s)+502g)->5C02(g)) delta H--1967.5 kJ 6H2(g)+36 2(g)--> 6H2Q(g)) delta H=-1450.5 kJ 5C(s)+6H2(g)>C512(1) Finally, the deltaHreaction is, deltaHreaction 3505.8 kJ-1967.5 kJ-1450.5 kJ-87.8 KJ

Answer 2

Calculate delta Hrxn for the following reaction:
5C(s)+6H2(g)--->C5H12(l)

RXN1 = C5H12(l)+8O2(g)---> 5CO2(g)+6H2O(g)      delta H= -3505.8 kJ
RXN2 = C(s)+O2(g)--->CO2(g)      delta H= -393.5 kJ
RXN3 = 2H2(g)+O2(g)---> 2H2O(g)                  delta H= -483.5 kJ

a) we need 5C(s) on the left, so multiply RXN 2 by 5

5C(s) + 5O2(g) = 5 CO2(g) H = 5(-393.5) = -1967.5 kJ

b) we need C5H12 on ht right side, so invert rxn 1

5CO2(g)+6H2O(g) ---> C5H12(l)+8O2(g) H= 3505.8 kJ

add both equations

5C(s) + 5O2(g) + 5CO2(g)+6H2O(g) ---> C5H12(l)+8O2(g)+5 CO2(g) H = -1967.5 +3505.8 = 1538.3

simplify

5C(s) + 5O2(g) + 6H2O(g) ---> C5H12(l)+8O2(g) H = 1538.3

we need 6H2 on the lef tside, so mult8iply RXN 3 by3

6H2(g)+3O2(g)---> 6H2O(g)                    delta H= 3*(-483.5) = -1450.5

add rxn

6H2(g)+3O2(g) + 5C(s) + 5O2(g) + 6H2O(g) ---> C5H12(l)+8O2(g)+6H2O(g)       H = 87.8

simplify

6H2(g)+5C(s) ---> C5H12(l) H = 87.8

order

5C(s) + 6H2(g) = C5H12(l) H = 87.8 kJ/mol

Answer 3

The fist step to solve this equation is to arrange the equations according to the place where you want them, for the first equation you have the C5H12 on the left side, you actually want it on the product side so the best thing to do is to invert the first equation

5CO₂(g)+6H₂O(g) ---> C₅H₁₂(l)+8O₂(g) $\Delta$H= 3505.8 kJ

You want 5 carbons on the left side, so you have to multiply by 5 the second equation, you have to multiply all the elements also you have to multiply by 5 the enthalpy provided

5 C(s)+ 5 O₂(g)---> 5 CO2(g) $\Delta$H= -1967.5 kJ

You want 6 H₂ in the overall equation, you have 2 H2 in equation 3, you have to multiply by 3 the equation to get 6H2 so:

6 H₂(g)+3 O₂(g)---> 6 H₂O(g) $\Delta$H= -1450.5 kJ

Now add the three equation

5CO₂(g)+6H₂O(g) ---> C₅H₁₂(l)+8O₂(g)

5 C(s)+ 5 O₂(g)---> 5 CO2(g)

6 H₂(g)+3 O₂(g)---> 6 H₂O(g)

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5 C + 6 H₂ + 8O₂ + 5CO₂ + 6H₂O  $\rightarrow$ C₅H_12(l)+ 8O₂ + 5CO₂ + 6H₂O

The underlined elements are repeated on both sides so you can remove it to make a net equation

5 C + 6 H₂$\rightarrow$ C₅H_12(l)

Our procedure is correct, similarly you have to add all the values to get the enthalpy you want to find:

3505.8 + (-1967.5) + (-1450.5) = 87.8 KJ / mol