How to use Hess's Law to calculate enthalpy for the reaction?

Calculate enthalpy for the reaction?

• I have a few problems I really need help on the first is: A) Use the (Delta)H^degree) data given to calculate (Delta)H^degree sign) for the reaction: C2H4(g) + H2(g) -> C2H4 Data: C2h4(g) + 3.5O2 -> 2CO2(g) + 3H2O(l) DeltaH^degree (KJ) -1560 C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l) DeltaH^degree (KJ) -1411 2H2(g) + O2 -> 2H2O(l) DeltaH^degree (KJ) -572 B)Using the Enthalpy of formation data calculate the DeltaH^degree subscript rxn for the reaction: 3NO2(g) + H2O(l) -> 2HNO3 (aq) + NO (g) C) Magnesium metal (0.200 mol) and a volume of aqueous hydrochloric acid that contains 5.00 mol of HCl are combined and react to completion. How many liters of hydrogen gas, measure in STP, are produced?

• Answer:

  1) first equation - flip it 2) second equation - leave untouched 3) third equation - divide by 2 Your target equation has a mistake, it should be C2H6 on the product side. C2H6 should also appear in the first data equation as a reactant. Here's the result of the above 3 steps: 2CO2(g) + 3H2O(l) ---> C2H6(g) + 3.5O2 DeltaH^degree (KJ) +1560 C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l) DeltaH^degree (KJ) -1411 H2(g) + (1/2)O2 -> H2O(l) DeltaH^degree (KJ) -286 When you add the three reactions together, the 3H2O cancels as does the 2CO2 and the 3.5O2. Get the answer by adding up the three enthalpies of the three revised equations just above. More Hess' Law here: http://www.chemteam.info/Thermochem/Thermochem.html One of the problem sets uses enthalpies of formation. For the third problem: Mg + 2HCl ---> MgCl2 + H2 The HCl is in excess (by far!), so concentrate on the Mg. The Mg:H2 molar ratio is 1:1, so 0.200 mole of H2 is produced. Since we are at STP, we can do this: 0.200 mol times 22.414 L/mol = 4.4828 L = 4.48 L (to three sig figs) If you wanted a volume at non-STP conditions, use PV = nRT.