(15 points) Given the following reactions: PH 3 (g) ® P(g) + 3H(g) DH rxn = 965

Question

(15 points) Given the following reactions:

PH3(g) ® P(g) + 3H(g)                          DHrxn = 965 kJ

O2(g) ® 2O(g)                                       DHrxn = 490. kJ

H2O(g) ® 2H(g) + O(g)                        DHrxn = 930. kJ

4P(g) + 5O2(g) ® P4O10(s)                    DHrxn = -6760. kJ

calculate the enthalpy change for the reaction

4PH3(g) + 8O2(g) 6H2O(g) + P4O10(s)

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(20 points)   Butane (C4H10(l)) is combusted in devices such as camping stoves. (a) Balance the equation for combustion of C4H10. (b) If 10.0g of C4H10(l) are combusted in an excess of oxygen, how many grams of water are formed? (c) If 10.0g C4H10(l) are combusted, what energy change accompanies the reaction?

         DHfo (C4H10(l)) = -147.6 kJ/mole

         DHfo (H2O(l))    = -285.8 kJ/mole

         DHfo (CO2(g))   = -393.5 kJ/mole

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Please answer above two questions, I do rate the answer, so please answer them seriously. Thank you!

Explanation / Answer

1)

Lets number the reaction as 1, 2, 3, 4, 5 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 5 = +4 * (reaction 1) +3 * (reaction 2) -6 * (reaction 3) +1 * (reaction 4)

So, deltaHo rxn for required reaction will be:

deltaHo rxn = +4 * deltaHo rxn(reaction 1) +3 * deltaHo rxn(reaction 2) -6 * deltaHo rxn(reaction 3) +1 * deltaHo rxn(reaction 4)

= +4 * (965.0) +3 * (490.0) -6 * (930.0) +1 * (-6760.0)

= -7010.0 KJ

Answer: -7010.0 KJ

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