The reaction below, carried out at 272.67 K and 1.00 bar, carried out at constant pressure...
The reaction below, carried out at 272.67 K and 1.00
bar, carried out at constant pressure was found to release 329 kJ
of heat and consume 21.8 g of oxygen.
Calculate the internal energy change associated with this
reaction.
2H2(g) + O2(g) → 2H2O(g)
a. 331 kJ
b. 329 kJ
c. -331 kJ
d. -327 kJ
e. -329 kJ
Homework Answers
According to 1st law of thermodynamics ;U = q + W
where U is internal
energy change
q is the heat release or gain to system
W is the work done
w=p*V =nRT as per ideal gas equation
n= no of moles = 21.8/32= 0.68125
W= 0.68125mole*272.67K* 8.314JK-1mol-1= 1544.37 J= 1.544kJ
U = -329kJ(
negative sign. for heat releasing) + 1.544kJ = -327.455kJ =
-327kJ
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