1. a) The combustion of pentane produces heat according to the equation: C 5 H 1
ID: 706713 • Letter: 1
Question
1.
a) The combustion of pentane produces heat according to the equation: C5H12 (l) + 8O2 (g) ? 5CO2 (g) + 6H2O (l) ?H°rxn = –3,510 kJ/mol How many grams of CO2 are produced per 2.50 × 103 kJ of heat released?
b) Ethanol (C2H5OH) burns according to the following equation: C2H5OH(l) + 3O2(g) ? 2CO2(g) + 3H2O(l) ?H°rxn = –1367 kJ/mol How much heat is released when 35.0 g of ethanol is burned?
c) Given the following ?H°rxn values:
H2(g) + ½ O2(g) ? H2O(l) ?H°f = –285.8 kJ/mol
H2O2(l) ? H2(g) + O2(g) ?H°rxn = 187.6 kJ/mol
Calculate the ?H°rxn for the reaction H2O2(l) ? H2O(l) + ½ O2(g)
d) Calculate the standard enthalpy of reaction (?H° RXN) for the following thermite reaction that involves aluminum and iron (III) oxide when one mole of aluminum is reacted.
2 Al (s) + Fe2O3 (s) ? Al2O3 (s) + 2Fe (l)
Note: ?H° f Al (s) = 0 kJ/mol; ?H° f Fe2O3 (s) = -822.2 kJ/mol;
?H° f Al2O3 (s)= -1669.8 kJ/mol; ?H° f Fe (l) = 12.40 kJ/mol
Explanation / Answer
1
a) C5H12 (l) + 8O2 (g) ? 5CO2 (g) + 6H2O (l)
1 mol C5H12 = 3510 kj
No of mol of C5H12 = (2.5*10^3)/3510 = 0.712 mol
nO of mol of CO2 produced = 0.712*5/1 = 3.56 mol
amount of CO2 produced = 3.56*44 = 156.64 g
b) 1 mol ethanol = –1367 kJ/mol
No of mol of ethanol = 35/46 = 0.761 mol
heat is released = 0.761*1367 = 1040.3 kj
c)
H2O2(l) ? H2(g) + O2(g) DH1 = 187.6 kJ/mol
H2(g) + ½ O2(g) ? H2O(l) DH2= –285.8 kJ/mol
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H2O2(l) ? H2O(l) + ½ O2(g) DH0 = DH1+DH2
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DH0 = 187.6-285.8 = -98.2 kj
D) 2 Al (s) + Fe2O3 (s) ? Al2O3 (s) + 2Fe (l)
DH0rxn = (1*DH0f,Al2O3 (s) + 2*DH0f,Fe (l)) - (2*DH0f, Al (s) + 1*DH0f,Fe2O3 (s))
= (1*-1669.8+2*12.4)-(2*0+1*-822.2)
= -822.8 kj
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