11)Balance each of the following reactions and then calculate the enthalpy of re
ID: 304042 • Letter: 1
Question
11)Balance each of the following reactions and then calculate the enthalpy of reaction for reaction 5 from the enthalpy of reactions using reactions (1) through (4):
(1) Fe (s) + O2 (g) ? FeO (s) ÄHrxn = -544 kJ
(2) Fe (s) + CO2 (g) ? Fe2O3 (s) + CO (g) ÄHrxn = 23 kJ
(3) FeO (s) + CO2 (g) ? Fe3O4 (s) + CO (g) ÄHrxn = 18 kJ
(4) CO (g) + O2 (g) ? CO2 (g) ÄHrxn = - 566 kJ
(5) Fe2O3 (s) + CO (g) ? CO2 (g) + Fe3O4 (s) ÄHrxn = ?
13)Methane, CH4, is commonly used in the laboratory as a fuel for Bunsen burners. The combustion reaction is fairly efficient in producing heat.
a)Write and balance the reaction for the combustion of methane.
b)Draw the Lewis structure for each reactant and product in the reaction.
c)Calculate the enthalpy of change using the bond energies
Bond
Dissociation Energy (kJ/mol)
Radius (pm)
H-H
432
74
H-C
411
109
H-O
459
96
C-C
346
154
C=C
602
134
C?C
835
120
C-O
358
143
C=O
799
120
O-O
142
148
O=O
494
121
d)Are the bonds stronger in the reactants or the products? Explain. Are the reactants or products more stable? Explain
16)Oxygen gas, generated by the reaction KClO3 (s) ? KCl (s) + O2 (g) is collected over water at 27°C in a 2.00 L vessel at a total pressure of 500 mm Hg. The vapor pressure of water at 27°C is 26.0 torr. How many moles of KClO3 were consumed in the reaction?
Bond
Dissociation Energy (kJ/mol)
Radius (pm)
H-H
432
74
H-C
411
109
H-O
459
96
C-C
346
154
C=C
602
134
C?C
835
120
C-O
358
143
C=O
799
120
O-O
142
148
O=O
494
121
Explanation / Answer
11.
To get equation (5) , do (1) - (2) + (3) + (4)
SO,
deltaH = deltaH1 - deltaH2 + deltaH3 + deltaH4
= - 544 - 23 + 18 - 566
= - 1115 kJ
13.
(a)
Balanced equation for the combustion of methane,
CH4 (g) + 2 O2 (g) ------------> CO2 (g) + 2 H2O (g)
(b)
(c)
deltaH = 4 * deltaHb(C-H) + 2 * deltaHb(O=O) - 2 * deltaHb(C=O) - 4 * deltaHb(O-H)
= 4 * ( 411 ) + 2 ( 494 ) - 2 ( 799 ) - 4 ( 459 )
= - 802 kJ
(d)
Since, deltaH = -ve, the bonds in poducts are stronger.
16.
Balanced equation,
2 KClO3 (s) -------> 2 KCl (s) + 3 O2 (g)
Temperature, T = 27 + 273 = 300 K
Pressure, P = 500 - 26.0 = 474.0 mmHg = 474.0 / 760. = 0.624 atm
Volume, V = 2.00 L
R = 0.0821 L.atm.K-1.mol-1
Ideal gas equation,
P V = n R T
n = 0.624 * 2.00 / ( 0.0821 * 300 )
n = 0.0507 mol
From the balanced equation ,
3 mol of O2 produced = 2 mol of KClO3 consumed
SO,
0.0507 mol of O2 produced = 2 * 0.0507 / 3 = 0.0338 mol of KClO3 consumed.
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