. 5 points - (a) Calculate the standard enthalpy of formation (?H o f ) of gaseo

Question

. 5 points - (a) Calculate the standard enthalpy of formation (?Hof) of gaseous diborane (B2H6)

2 B (s) + 3 H2(g) ---> B2H6(g)

using the following thermochemical information:

4 B(s) + 3 O2(g) ---> 2 B2O3(s) DHo = -2509.1 kJ

2 H2(g) + O2(g) ---> 2 H2O(l) DHo = -571.7 kJ

B2H6(g) + 3 O2(g) ---> B2O3(s) + 3 H2O(l) DHo = -2147.5 kJ

5 points - (b) Pentaborane (B5H9) is another boron hydride. What experiment(s) would you need to perform to yield the data necessary to calculate the heat of formation of B5H9(l).

Explain by writing out and summing any applicable chemical reactions.

2. Gasoline is composed of primarily hydrocarbons, including many with eight carbon atoms (called octanes). One of the cleanest burning octanes is a compound called 2,3,4-trimethylpentane - C8H18, which has the following structural formula:

CH3CH(CH3)CH(CH3)CH(CH3)CH3

The complete combustion of one mole of this compound to CO2(g) and H2O(g) leads to ?Ho = -5069 kJ.

2 points - (a) Write a balanced equation for the combustion of 1 mol of C8H18(l);

2 points - (b) Write a balanced equation for the formation of C8H18(l) from its elements;

6 points - (c) By using the information above and the standard heats of formation of H2O(g) = -241.8 kJ/mole and CO2(g) = -393.5 kJ/mole, calculate ?Hof for 2,3,4-trimethylpentane.

Explanation / Answer

a) The equation for the formation of B2H6 is:

2 B(s) + 3 H2(g) ---> B2H6(g)

To get this from the three equations above (call them A, B and C in that order) you need to halve A, add it to 3/2 of B and then subtract C. You will find everything else cancels out and you are left with the equation you want.

So,

?H = (0.5 * -2509.1) + (1.5 * -571.7) - (-2147.5)

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