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 , ∆H^o , is -137.0 kJ for the chemical reaction C₂H_4(g) + H_2(g) → C₂H_6(g) How much heat is liberated when 400.0 g of C₂H₆ is produced ?

c) Given the following thermochemical reactions:

CH_{4 (g)} + 1/2 O_2(g) → CH₃OH _(l) ∆H^o = -163.9 kJ

CO_{2 (g)} + 2H₂O_(l) → CH_{4 (g)} + 2 O_{2 (g)} ∆H^o = 890.4 kJ

What is the standard enthalpy of formation of CO_2(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.

Answer 1

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)