Chemistry Equilibrium Shift and Energy Calculations

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
1mol =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 = 0.9352mol × 890.4 kJ
1mol =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 °C
Q=mCm ∆ T∆ T = Q
mCm
= 128 J
10.5 g × 0.449 J
g °C
27.15 °C Fina Temperature=25+27.15=52.15 °C
15. The molar heat capacity of ethanol, C2 H5 O given the specific heat capacity
CC2 H 5 OH =2.46 j
g °C M C2 H 5 O=2.46 j
g °C × 46.07 g
mol ¿ 113.332 J
mol °C
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 °C × ( 100−x ) =150 g ×4.184 J
g °C ( x−20 )
4515−45.15 x=627.6 x−12552 672.75 x=17067 x=25.37 °C
17. Burning sulfur in excess oxygen
Mass of sulfur =4.05
Ns = 4.05
32.065 =0.1263Q=1000 g × 4.184 J
g °C × 8.88 °C¿ 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 °C ×6.9 °C=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 °C ( 24.83−23.00 )
24.6 g × ( 110−24.83 )
¿ 0.4568 J
g °C
The equilibrium expression for the following is:
8. A s4 O6 ( s ) +6 C ( s ) ⇔ A s4 ( g ) +6 C O ( g )