1) Combustion of 7.21 g of liquid benzene (C6H6) causes a temperature rise of 50
ID: 476636 • Letter: 1
Question
1) Combustion of 7.21 g of liquid benzene (C6H6) causes a temperature rise of 50.3 C in a constant-pressure calorimeter that has a heat capacity of 5.99 kJ/C. What is delta H for the following reaction?
C6H6 (l) + 15/2 O2 (g) 6CO2(g) + 3H2O(l)
A. 302 kJ/mol-rxn
B. 41.8 kJ/mol-rxn
C. -3.27x10^3 kJ/mol-rxn
D. -302 kJ/mol-rxn
E. -41.8 kJ/mol-rxn
2) CaO(s) reacts with water to form Ca(OH)2 (aq). If 6.50 g CaO is combined with 99.70 g H2O in a coffee cup calorimeter, the temperature of the resulting solution increases from 21,.7 C to 43.1 C . Calculate the enthalpy change for the reaction per mole of CaO. Assume specific heat capacity of solution is 4.18 J/g*k
A. -165 kJ/mol
B. -532 kJ.mol
C. -82.0 kJ/mol
D. -9.42 kJ/mol
E. -1.45 kJ/mol
Explanation / Answer
The heat realeased per mole of benzene is Q
-nQ=Calaorimeter constant *temperature rise
-nQ=Cv*delta T
nQ=-(5.99 kJ/C*50.3 C)=-301.297 kJ
Moles of benzene=Mass of benzene/molecualr weight=7.21 g/78 g/mol=0.0921 mol
the heat released Q=-301.297 kJ/0.0921 mol=-3271.41 kJ/mol=-3.27x10^3 kJ/mol-rxn
So, option C is the correct answer.
Post the other question separately
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