Part A For which of the following reactions is H rxn equal to H f of the product
ID: 535855 • Letter: P
Question
Part A
For which of the following reactions is Hrxn equal to Hf of the product(s)?
You do not need to look up any values to answer this question.
Check all that apply.
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Incorrect; Try Again; 4 attempts remaining
Look again at the coefficients in the reactions. The heat of formation is defined by a reaction that forms one mole of product.
Part B
The combustion of propane, C3H8, occurs via the reaction
C3H8(g)+5O2(g)3CO2(g)+4H2O(g)
with heat of formation values given by the following table:
Calculate the enthalpy for the combustion of 1 mole of propane.
Express your answer to four significant figures and include the appropriate units.
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Part C
What is Hrxn for the following chemical reaction?
CO2(g)+2KOH(s)H2O(g)+K2CO3(s)
You can use the following table of standard heats of formation (Hf) to calculate the enthalpy of the given reaction.
Express the standard enthalpy of reaction to three significant figures and include the appropriate units.
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Part D
Calculate the mass of glucose metabolized by a 69.5 kg person in climbing a mountain with an elevation gain of 1330 m . Assume that the work performed in the climb is four times that required to simply lift 69.5 kg by 1330 m .
Express your answer to three significant figures and include the appropriate units.
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Na(s)+12F2(l)NaF(s) S(s)+O2(g)SO2(g) 2Na(s)+F2(g)2NaF(s) SO3(g)12O2(g)+SO2(g) Na(s)+12F2(g)NaF(s) SO(g)+12O2(g)SO2(g)Explanation / Answer
1) Hf products - Hf reactants = Hrxn
In order for Hrxn to equal the Hf of the products,the Hf of the reactants must be zero. In order for that to happen, all the reactants must be in their elemental state at ambient temperature.
2, 3, 4, 5 are correct
2)C3H8(g)+5O2(g)3CO2(g)+4H2O(g)
Sum Hf products - Sum Hf reactants
[3(-393.5) + 4(-241.8)] - (-104.7) + 5(0)
(-1180.5) + (-967.2) - (-104.7) + 0
-2147.7 + 104.7
Hcombution = - 2043 kJ/mole (endothermic)
3) CO2(g)+2KOH(s)H2O(g)+K2CO3(s)
CO2 = 393.5 kJ
KOH = 424.7
H20= -241.8 kJ
K2CO3 = -1150 kJ
Sum Hf products - Sum Hf reactants
{(-241.8 )+ (-1150)}-{(-393.5)-2(-424.7)}
=-1391.8+1242.9
= -148.9 kJ
4.)