University Chemistry Assignment: Thermodynamics and Equilibrium

This document presents a comprehensive solution to a chemistry assignment, addressing key concepts in thermodynamics and chemical equilibrium. It begins with calculations involving enthalpy changes, determining energy released or absorbed during the combustion of methane and related reactions, including calculations for energy released based on moles and mass of reactants. The assignment then delves into heat capacity calculations, determining the heat capacity of water and calculating final temperatures in scenarios involving heat transfer between substances. Furthermore, the document explores equilibrium principles, calculating equilibrium constants (Keq) and reaction quotients (Q) to predict the direction of reaction shifts under various conditions. The assignment also includes a detailed analysis of Le Chatelier's principle, predicting the effects of changes in concentration, temperature, and pressure on equilibrium position and reactant/product concentrations for several chemical reactions. The solutions are presented with step-by-step calculations and explanations, providing a thorough understanding of the concepts involved.

Chemistry
1 Given that when 1 mole of methane burns in air releases 890.4 kJ of energy according to the
equation below:
C H 4 ( g ) +2O2 ( g ) →C O2 ( g ) +2 H2 O ( l ) ∆ H=−890.4 kJ
a) Energy given out when 2.0 moles of gas are burned is given by:
If 1 mole =-890.4 kJ
Then, 2 moles will be
2.0 mol ×890.4 kJ
1 mol =1780.8 kJ
b) Energy released when 22.4g of the gas is burned
No .moles of C H4 ( g )= mass
molar mass = 22.4 g
16.04 g
mol
=1.3965 mol
∴ energy released= 1.3965 mol ×890.4 kJ
1 mol =1243.45 kJ
c) To get 15 g of C H 4 (g ) ,one needs
No .moles of C H4 ( g )= mass
molar mass = 15 g
16.04 g
mol
=0.9352 mol
∴ energy absorbed= 0.9352 mol ×890.4 kJ
1 mol =832.67 kJ
13. When 1.5 ×103 J of heat energy is absorbed by a beaker of water. The change in
temperature is 3.1 ˚C. The heat capacity of the beaker of water can be determined as:
Q=c ∆T c= Q
∆ T = 1.5× 103
3.1 =483.87 j
K
14. If 10.5 g of iron at 25 ˚C, absorbs 128 J of heat, the final temperature of the metal will be:

Given Cm=0.449 J
g ℃
Q=mCm ∆ T∆ T = Q
mCm
= 128 J
10.5 g × 0.449 J
g ℃
27.15 ℃ Fina Temperature=25+27.15=52.15℃
15. The molar heat capacity of ethanol, C2 H5 O given the specific heat capacity
CC2 H 5 OH =2.46 j
g ℃ M C2 H 5 O=2.46 j
g ℃ × 46.07 g
mol ¿ 113.332 J
mol ℃
16. Initial temperature of aluminum is 100 ˚C, Initial temperature of water is 20 ˚C. Here, the
heat released by the metal equal to heat absorbed by water. Let the final temperature be x
and assuming no heat loss to the surrounding:
mm cm ( 100−x ) =mw cw ( x−20 )50.0 g× 0.903 J
g ℃ × ( 100−x ) =150 g × 4.184 J
g ℃ ( x−20 )
4515−45.15 x=627.6 x−12552672.75 x=17067 x=25.37 ℃
17. Burning sulfur in excess oxygen
Mass of sulfur =4.05
Ns = 4.05
32.065 =0.1263Q=1000 g × 4.184 J
g ℃ ×8.88 ℃ ¿ 37153.92 J0.1263 produces37.15392 kJ
1 mol=37.15392 kJ
0.1263 mol =294.1581345 kJ
mol

18. The amount of energy released when 1 kg of water freezes mass =1000 g, , cw=6.02 kJ
mol
No of moles of HO= 1000
18.02 g
mol
=55.493896 mol
Energy released =55.493896 mol ×6.02 kJ
mol ¿ 334.0733 kJ
19. Using the coffee calorimeter
Given
vHCl =150 mL , M HCl =1.00 mol
L vNaOH =150 mL , M NaOH=1.00 mol
L , ∆ T =29.5−22.6=6.9
NaO H ( aq ) +HC l (aq ) → NaC l (aq )+ H2 O (l ) q=mHCl+ NaOh × cwater × ∆ T
¿ 300 g× 4.184 J
g ℃ × 6.9℃=8656.74 J
20. Initial temperature of Nickel is 110 ˚C, Initial temperature of water is 23.0 ˚C. Here, the
heat released by the metal equal to heat absorbed by water. The final temperatire of water
and Nickel is24.83˚C.assuming no heat loss to the surrounding
m¿ c¿ ∆ T ¿=mw cw ∆ T w c¿= mw cw ∆ T w
m¿ ∆ T¿ c¿=
125 g × 4.184 J
g ℃ ( 24.83−23.00 )
24.6 g × ( 110−24.83 )
¿ 0.4568 J
g ℃
The equilibrium expression for the following is:
8. A s4 O6 ( s ) +6 C (s ) ⇔ A s4 ( g ) +6 C O ( g )
9. SnO2 ( s ) +2 C O ( g ) ⇔ S n ( s ) +2 C O2 ( g )
10. CaC O3 (s ) ⇔ CaO ( s ) +C O2 ( s )

11. for the reaction 2 C O ( g ) ⇔C ( s ) +C O2 ( s ) Keq=7.7 ×10−15. At a perticlar time the concentration
were measure [ CO ] =0.034 M , [ C O2 ] =3.6× 10−17 M,
Q= [ 3.6 × 10−17
[ 0.034 ]2 ]=3.1142 ×10−14 Q> Keq
The reaction is not at equilibrium. The reaction will proceed to the reactant (to the left)
12. For the reaction
N2 O4 ( g ) ⇔2 N O2 ( g ) keq=0.2 [ N2 O4 ] =2.0 M , [ N O2 ] =0.2 MQ= [ 0.2 ]2
[2.0 ] 0.020.02<0.2 Q< Keq
The reaction is not at equilibrium, the reaction will proceed the in the right
N2 O4 ( g ) ←2 N O2 ( g )
13. Given the reaction
2 ICl ( g ) ⇔ I2 ( g ) +C l2 ( g ) keq=0.11 [ ICl ] =2.5 M , [ I2 ] =2.0 M ∧ [ C l2 ] =1.2 M Q= [ 2.0 ] [1.2 ]
[ 2.5 ] =0.96
0.96> 0.11Q> K eq
The reaction is not at equilibrium. The reaction will proceed to the left (towards the
reactants)
14. At 340 ˚C, keq =0.064 for the reaction Fe2O3(s) +3H2(g) 2Fe(s) +3H2O(g). Given
that [H2]=0.45 M and [H2O]= 0.37 M.
Q= [ 0.37 ]3
[ 0.45 ]
3 =0.55586Q>Keq
The reaction need to proceed towards the reactant (that is to the left)

Completing the given chart, by filling left, right or none for the equilibrium shift and decreases,
increases or remains the same for the concentration of the reactant and the produces and for the
value K
N2 ( g ) + 3 H 2 ( g ) ⇔2 N H 3 ( g ) +22.0 kcal
Stress Equilibri
um shift
[N2] [H2} [NH3] K
18. Add N2 Right ________
_
Decreases Increases Remain
the same
19. Add H2 Right Decrease ________ Increases Remain
the same
20. Add NH3 Left Increases Increases ______ Remain
the same
21. Remove N2 Left ________ Increases Decrease Remain
the same
22. Remove H2 Left Increases ________ Decrease Remain
the same
23. Remove NH3 Right Decrease Decrease _______ Remain
the same
24. increase
Temperature
Left Increases Increase Decrease Decrease
25.decrease Right Decrease Decrease Increases Increase

Temperature
26. increase Pressure Right Decrease Decrease Increases Increase
27.decrease Pressure Left Increases Increase Decrease Decrease