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Enthalpy of 4B+3O2-->2B2O3

  1. Oct 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the enthalpy of the reaction

    4B(s)+3O2(g)→2B2O3(s)

    given the following pertinent information:
    1. B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035kJ
    2. 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36kJ
    3. H2(g)+1/2O2(g)→H2O(l), ΔH∘C=−285kJ
    4. H2O(l)→H2O(g), ΔH∘D=+44kJ

    2. Relevant equations

    ΔH=Sum of all ΔH of each reaction. (Hess' Law?)

    3. The attempt at a solution

    2[2B(s)+3H2(g)→B2H6(g)] ΔH=72 J
    2[3O2(g)+B2H6(g)→B2O3(s)+3H2O(g)] ΔH=-4070kJ

    ΔH=3998 kJ


    This incorporates the correct reactants and products. We still need to add H2, but there's no way to react only the given reactants, B and O2, into H2 at all. Also, there is no way to react out the H2O from the given equations.

    The actual solution

    6[H2O(g)→H2O(l)] ΔH=-264 kJ
    6[H2O(l)→H2(g)+1/2O2(g)] ΔH=1710J
    2[2B(s)+3H2(g)→B2H6(g)] ΔH=72 kJ
    2[3O2(g)+B2H6(g)→B2O3(s)+3H2O(g)] ΔH=-4070kJ

    ΔH=-2552

    Colours correspond to cancelled products.

    I understand the process given, but not how it relates to the goal equation. This process requires water to be added as a reactant and for 3H2 and 3H2O to be products, which are not included in the goal equation. Adding the extra reactions with water do not seem directly part of the goal equation.
     
  2. jcsd
  3. Oct 30, 2015 #2

    Borek

    User Avatar

    Staff: Mentor

    But is necessary to cancel out hydrogen and water. Your solution is incorrect as your equation is not identical with the goal equation, and it must be.
     
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