please answer the following enthalpie questions, 1. calculate the standard entha
ID: 1003772 • Letter: P
Question
please answer the following enthalpie questions,
1. calculate the standard enthalpy change, H°rxn, in kJ for the following chemical equation, using only the thermochemical equations below:
NO(g) + NO3(g) 2NO2(g)
Report your answer to three significant figures in scientific notation.
2. a)The enthalpy of combustion (H°c) of cyclopentanone (C5H8O) is -2873.50 kJ/mol. Using the appropriate information given below, calculate the enthalpy of formation (H°f), in kJ/mol, for cyclopentanone.
H°f (CO2 (g)) = -393.51 kJ/mol
H°f (H2O (l)) = -285.83 kJ/mol
2.b) Determine the mass (in g) of cyclopentanone produced, if H° was determined to be -182.77 kJ during an experiment in which cyclopentanone was formed.
3.1. Using the enthalpies of formation given below, calculate H°rxn in kJ, for the following reaction.
Report your answer to two decimal places in standard notation.
2Na(s) + 2H2O(l) 2NaOH(s) + H2(g)
Na (s): 0.00 kJ/mol
H2O (l): -285.83 kJ/mol
NaOH (s): -425.93 kJ/mol
H2 (g): 0.00 kJ/mol
3b) Calculate the amount of heat absorbed/released (in kJ) when 20.1 grams of NaOH are produced via the above reaction.
Report your answer to two decimal places, and use appropriate signs to indicate heat flow direction.
Explanation / Answer
Answer – 1) We are given chemical equations with given standard enthalpy change, Horxn and we need to calculate the standard enthalpy change for the following reaction –
NO(g) + NO3(g) ----> 2NO2(g) , Horxn = ?
NO(g) + ½ O2(g) ----> NO2(g) , Horxn = -56.5 kJ …..1
½ N2(g) + O2(g) ----> NO2(g) , Horxn = 33.8 kJ …..2
N2(g) + 3O2(g) -----> 2NO3(g). Horxn = 142.3 kJ …..3
Now using the Hess’s law we need to arrange this equations in such way when we added these equation we need to get the final equation is which is we need Horxn .
We need NO2 in product aide and NO and NO3 in reactant side
We need to dived equation number 3 by 2 and then reverse and added all these reactions as follow –
NO(g) + ½ O2(g) ----> NO2(g) , Horxn = -56.5 kJ
½ N2(g) + O2(g) ----> NO2(g) , Horxn = 33.8 kJ
NO3(g) ----> ½ N2(g) + 3/2 O2(g) Horxn = -71.15 kJ
NO(g) + NO3(g) ----> 2NO2(g) , Horxn = -9.39*101 kJ
2) a) We are given, the enthalpy of combustion (H°rxn) of cyclopentanone (C5H8O) is -2873.50 kJ/mol.
H°f CO2(g) = -393.51 , H°f = H2O(l) = -285.83 kJ
We know the balanced combustion reaction -
2 C5H8O(g) +13 O2(g) ---> 10 CO2(g) + 8 H2O(l), H°rxn = -2873.50 kJ/mol
Horxn = Hof of product – Hof of reactant
Horxn = [ (10* Hof CO2(g)) + (8* Hof H2O(l) )] – [ 2* Hof C5H8O(g) + 13* Hof O2(g)]
-2873.50 kJ = (10*-393.5 kJ) + (8* -285.83) ] – (2* Hof C5H8O(g) +(13*0.00 kJ)]
-2873.50 kJ = -6221.6 - 2* Hof C5H8O (g)
-2* Hof C5H8O (g) = -2873.50 kJ +6221.6 kJ
= -3348.1 kJ
Hof C5H8O (g) = -3348.1kJ / 2
= - 1674.05 kJ/mol
The standard enthalpy of formation of H C5H8O 2O(g) is - 1674.05 kJ/mol.
b) We are given the , Ho = -182.77 kJ and we know,
Hof C5H8O (g) = -1674.05 kJ/mol
So,
-1674.05 kJ = 1 moles of C5H8O
So, -182.77 kJ = ?
= 0.109 moles of C5H8O
We know,
Mass of C5H8O = moles * molar mass
= 0.109 moles * 84.12 g/mol
= 9.18 g
3.a) We are given the standard enthalpy of compound and need to calculate the standard enthalpy of reaction as follow –
2Na(s) + 2H2O(l) -----> 2NaOH(s) + H2(g)
Na (s): 0.00 kJ/mol
H2O (l): -285.83 kJ/mol
NaOH (s): -425.93 kJ/mol
H2 (g): 0.00 kJ/mol
We know,
Horxn = Hof of product – Hof of reactant
Horxn = [ (2* Hof NaOH(s)) + (Hof H2(g) )] – [ 2* Hof Na(s) + 2* Hof H2O(l)]
= ( 2*-425.93 + 0.00 kJ) – ( 2*0.00 kJ + 2*-285.83 kJ)
= -280.2 kJ/mol
2Na(s) + 2H2O(l) -----> 2NaOH(s) + H2(g), Horxn = -280.2 kJ/mol
b) We are given, mass of NaOH = 20.1 g
we need to calculate the moles of NaOH
we know,
moles of NaOH = 20.1 g / 40.0 g.mol-1
= 0.503 moles
We know,
2 moles of NaOH = -280.2 kJ
So, 0.503 moles of NaOH = ?
= -70.4 kJ
So, -70.4 kJ heat released (in kJ) when 20.1 grams of NaOH are produced via the above reaction.
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