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ID: 695992 • Letter: F
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4) What happened to the Hindenbere? Nobody seems to know. Was it a spark? Sabotage? Lightening? Either way it did burn, a lot. We do know it was filled with hydrogen gas. The Zeppelin was filled with 2.0 x 108 Lof H. Show all work for the below. a moll below using the heat of formation for the reactants and Calculate the enthalpy of the reaction(kJ/ products from appendix C from your text book b) Hydrogen gas and oxygen gas can also produce liquid hydrogen peroxide(H 02) according to the following equation: Calculate the enthalpy of the reaction(k/ products from appendix C from your text book Using Hess's Law and the chemical equations and enthalpies from parts a) and b), calculate the enthalpy for the decomposition of hydrogen peroxide to liquid water and oxygen gas. Show all work as to how you applied Hess's law to this problem. f the 2.0 x 108 L of hydrogen gas corresponds to 8.32 x 10% mol of H2, how much heat was exchanged upon the burning of this H2 gas? Comment on the magnitude of this number. c) d)Explanation / Answer
a)
HRXN = Hprod - Hreact
HRXN = 2*H2O - (2H2 + O2)
HRXN = 2*-285.8 - 2*0 - 0
HRXN = -571.6 kJ/mol
b)
H2 + O2 = H2O2
HRXN = H2O2 - (H2 + O2)
HRXN = -187.8 - (0+0) = -187.8 kJ/mol
c
enthalpy for
H2O2 = H2O + O2
given
2H2 + O2 = 2H2O H = -571.6
H2 + O2 = H2O2 H = -187.8
invert rxn 2
H2O2 = H2 + O2 H = 187.8
divide rxn 1 by 2
H2 + 1/2O2 = H2O H = -571.6 /2 = -285.8
add both
H2 + 1/2O2 + H2O2 = H2 + O2 + H2O H = 187.8-285.8
simplify
H2O2 = 1/2O2 + H2O H = -98 kJ/mol
d.
mol of H2 = 8.32*10^6
from rxn 1, w eknow that HRXN -571.6 kJ/mol
Qtotal = (571.6/2)(8.32*10^6) = 2377856000 kJ release, which is a lot for moles of H2
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