a) Find q, w , E when a system gains 10.0 kJ of heat from the surroundings and d
ID: 550957 • Letter: A
Question
a) Find q, w , E when a system gains 10.0 kJ of heat from the surroundings and does 14.0 KJ of work on the surroundings.
b) If the standard enthalpy of reaction , Ho , is -137.0 kJ for the chemical reaction C2H4(g) + H2(g) C2H6(g) How much heat is liberated when 400.0 g of C2H6 is produced ?
c) Given the following thermochemical reactions:
CH4 (g) + 1/2 O2(g) CH3OH (l) Ho = -163.9 kJ
CO2 (g) + 2H2O(l) CH4 (g) + 2 O2 (g) Ho = 890.4 kJ
What is the standard enthalpy of formation of CO2(g) , if that for CH3OH(l) is -238.7 kJ and -285.8 kJ for liquid water ? Hint: Use Hess’s Law.
Step by step explanation please.
Explanation / Answer
a. As per first law of thermodynamics
E = q-w
given that q = energy gained = 10 kJ
work done, w = 14 kJ
then
E = 10-14 = -4 kJ
b.
Ho = -137.0 kJ (this is the value for one mole)
mass of ethane produced = 400 g
molar mass of ethane = 30 g/mol
number of moles of ethane =400/30 = 13.33 moles
H reaction = 13.33 x -137.0 kJ = -1826.66 kJ
omitting the sign (it is shoing that energy is released)
energy released = 1826.66 kJ
c).
CH4 (g) + 1/2 O2(g) CH3OH (l) Ho = -163.9 kJ Equation 1
CO2 (g) + 2H2O(l) CH4 (g) + 2 O2 (g) Ho = 890.4 kJ equation 2
H2 (g)+ 1/2 O2 (g) H2O (l) Ho = -285.8 kJ Equation 3
C(s) + 2 H2 (g) + 1/2 O2 (g) CH3OH (l) Ho = -238.7 kJ Equation 4
Reverse equation 1 and 2 (while reversining sign of Ho, will change)
CH3OH (l) CH4 (g) + 1/2 O2(g) Ho = +163.9 kJ equation 5
CH4 (g) + 2 O2 (g) CO2 (g) + 2H2O(l) Ho = -890.4 kJ equation 6
multiply equation 3 by 2 and reverse it
2H2O(l) 2H2 (g)+ O2 (g) Ho = 571.6 kJ Equation 7
add equation 4,5 ,6 and 7
C(s) + 2 H2 (g) + 1/2 O2 (g) + CH3OH (l) +CH4 (g) + 2 O2 (g) +2H2O(l) CH3OH + CH4 (g) + 1/2 O2(g)+CO2 (g) + 2H2O(l) +2H2 (g)+ O2 (g)
similar terms calncel each other
C(s) +O2 (g) CO2 (g)
H = -238. 7 + +163.9 + -890.4+571.6 = -393.6 kJ (this is the formation enthalpy of CO2)
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