1. Copper(I) oxide is converted to copper(II) oxide according to the following e
ID: 918992 • Letter: 1
Question
1. Copper(I) oxide is converted to copper(II) oxide according to the following equation:
Cu2O(s) + ½O2(g) 2 CuO(s)
with H° = -146 kJ.
Given that Hf° for Cu2O(s) = -168.6 kJ/mol, what is Hf° for CuO(s)?
2. Calculate the mass of methane that must be burned to provide enough heat to convert 390.0 g of water at 26.0°C into steam at 122.0°C. (Assume that the H2O produced in the combustion reaction is steam rather than liquid water.)
3. Compute the standard free energy change for the following reaction.
O3(g) + NO(g) NO2(g) + O2(g)
Explanation / Answer
Answer – 1) We are given, reaction –
Cu2O(s) + ½O2(g) ------> 2 CuO(s) H° = -146 kJ
Hf° for Cu2O(s) = -168.6 kJ/mol , Hf° for CuO(s) = ?
We know,
H°rxn = sum of the Hf° for product - sum of the Hf° for reactant
-146 kJ = [2* Hf° CuO(s)] – [Hf° Cu2O(s) + ½ *Hf° O2]
-146 kJ = 2* Hf° CuO(s) – (-168.6 kJ – ½ 0.00)
-146 kJ = 2* Hf° CuO(s) + 168.6 kJ
-146 kJ -168.6 kJ = 2* Hf° CuO(s)
So, 2*Hf° CuO(s) = -314.6
So, Hf° CuO(s) = -314.6 kJ / 2
= -157.3 kJ
2) Given, mass of water = 390.0 g , ti = 26.0oC, tf = 122.0oC
We need to calculate the heat from 26oC to 100oC
So, heat q1 = m*C*t
= 390 g * 4.184 J/goC * (100-26)oC
= 120750.2 J
Heat from 100oC to 100oC
We know,
Heat, q2 = Hvap * m
= 2260 J/g * 390 g
= 881400 J
Heat from 100oC to 120oC
Heat, q3 = m * C * t
= 390 g * 2.010 J/goC * (120-100)oC
= 15678 J
So total heat q = q1+ q2+ q3
= 120750.2 J + 881400 J +15678 J
= 1.018*106 J
We know reaction,
CH4 + 2 O2 -----> CO2 + 2 H2O ,
Ho = -1.018*106 J
So, Ho rxn = sum of the Hf° for product - sum of the Hf° for reactant
= [Hf° CO2 + 2*Hf° H2O] – [Hf° CH4 + 2*Hf° O2]
= (-393.51 + 2*-241.82) – (-74.85 + 2*0.0)
= -802.3 J
So, for 1 moles of CH4 = -802.3 J
So, ? = -1.018*106 J
= - 1.018*106 J * 1 moles of CH4 / -802.3 J
= 1268.6 moles
Mass of CH4 = 1268.6 moles * 16.0426 g/mol
= 20352.3 g of CH4
3) Given, O3(g) + NO(g) -----> NO2(g) + O2(g)
G°rxn = sum of the Gf° for product - sum of the Gf° for reactant
= [ Gf° NO2(g) + Gf° O2] – [Gf° O3(g) +Gf° NO]
= (51.30 + 0.0) – ( 163.18 + 86.57)
= -198.4 kJ
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