Calculate H rxn for the followingreactions: a) Fe 2 O 3 (s) + 3CO(g) ---> 2Fe(s)

Question

Calculate Hrxn for the followingreactions: a) Fe2O3(s) + 3CO(g) ---> 2Fe(s) +3CO2(g) Use the following reactions and given H's. 2Fe(s) + 3/2O2(g) --->Fe2O3(s)    H =-824.2kJ CO(g) + 1/2O2(g) --->CO2(g)    H = -282.7kJ b) 5C(s) + 6H2(g) --->C5H12(l) Use the following reactions and given H's. C5H12(l) + 8O2(g) --->5CO2(g) + 6H2O(g)    H =-3505.8kJ C(s) + O2(g) ---> CO2(g)  H = -393.5kJ 2H2(g) + O2(g) --->2H2O(g)    H = -483.5kJ Calculate Hrxn for the followingreactions: a) Fe2O3(s) + 3CO(g) ---> 2Fe(s) +3CO2(g) Use the following reactions and given H's. 2Fe(s) + 3/2O2(g) --->Fe2O3(s)    H =-824.2kJ CO(g) + 1/2O2(g) --->CO2(g)    H = -282.7kJ b) 5C(s) + 6H2(g) --->C5H12(l) Use the following reactions and given H's. C5H12(l) + 8O2(g) --->5CO2(g) + 6H2O(g)    H =-3505.8kJ C(s) + O2(g) ---> CO2(g)  H = -393.5kJ 2H2(g) + O2(g) --->2H2O(g)    H = -483.5kJ

Explanation / Answer

To solve these problems, you will try to manipulate the givenreactions by multiplying or dividing coefficients in thereaction or by inverting the reactions to get to your targetreaction.

a) Your target reaction:

Fe2O3(s) + 3CO(g) ---> 2Fe(s) +3CO2(g)

You want Fe2O3(s) on the left side of thereaction, so invert the first reaction equation to get:

Fe2O3 (s) ---> 2Fe+3/2O2.

Everything you do to the reaction, you do to the enthalpy;invert the H to get: +824.2 kJ.

There is a CO(g) + 1/2O2(g) --->CO2(g)   equation. The CO is on theleft side and the CO2 is on the right side, as needed, but you haveto multiply all reactants by 3 to get the correct coefficient:

3CO(g) + 3/2O2(g) --->3CO2(g)               

Since you multiplied the reaction by 3, multiply the H by3 to get: 3(-283.7 kJ)=-851.1 kJ.

You now have the following equations:

Fe2O3(s)                              --->2Fe     +   3/2O2(g)      H = +824.2 kJ

              3CO(g) + 3/2O2(g) --->3CO2(g)                       H = - 851.1 kJ

Now you can cancel out terms that appear on both sides to getyour target equation, and then add the enthalpies to get theH.

   Fe2O3(s)                              --->2Fe     +   3/2O2(g)      H = +824.2kJ

                   3CO(g)+ 3/2O2(g) --->3CO2(g)                       H= - 851.1 kJ

Fe2O3 (s) + 3CO(g)              ---> 2Fe (s) +3CO2            H= +824.2-851.1kJ  

b)   Your target reaction:

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

Manipulation:

The basic procedure is the same as before. You will want toinvert the equations to make sure the terms appear on the same sideas the target reaction, making sure to invert the enthalpiesaccordingly. If any of the coefficients are different, youwill also want to multiply or divide them to get the targetreaction coefficients. Again, you have to multiply or dividethe enthalpies by the same number!

Once you have the correct reactions, cancel out terms thatappear on both sides and add the enthalpies to get yourH.

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

         6H2  +    3O2        --->   6H2O                       H= 3(-483.5) kJ

   5CO2    +     6H2O   --->   C5H12    +   8O2          H= +3505.8 kJ

5C + 6H2  + 8O2 + 5CO2  +  6H2O  --->5CO2 +C5H12 + 6H2O + 8O2

5C + 6H2 --->C5H12                   H = 5(-393.5) + 3(-483.5) +3505.8 kJ

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