a In the reaction of two moles of gaseous sulfur dioxide and one mole of gaseous oxygen to form two moles of gaseous sulfur trioxide, two moles of products are formed from 3 moles of reactants. If this reaction is done at 1.00 atm pressure (and at 0 °C), the volume is reduced by 22.4 L. In this reaction, how much work is done on the system ( SO2, O2, SO3) by the surroundings? Work 2269.72 Calculate the amount of work done on the surroundings using the equation w = -P × AV. w = −101.3 J/(L · atm) × 1.00 atm × (-22.4 L) A,U° : Correct = J - b The enthalpy change for this reaction is -197.9 kJ. Use this value, along with the answer to previous part, to calculate AU, the change in internal energy in the system. kJ

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a In the reaction of two moles of gaseous sulfur dioxide and
one mole of gaseous oxygen to form two moles of gaseous
sulfur trioxide, two moles of products are formed from 3
moles of reactants. If this reaction is done at 1.00 atm
pressure (and at 0 °C), the volume is reduced by 22.4 L. In
this reaction, how much work is done on the system (
SO2, O2, SO3) by the surroundings?
Work 2269.72
Correct
Calculate the amount of work done on the surroundings
using the equation w =
-P × AV.
J
-
=
w = −101.3 J/(L · atm) × 1.00 atm × ( −22.4 L)
E
b The enthalpy change for this reaction is -197.9 kJ. Use this
value, along with the answer to previous part, to calculate
AU, the change in internal energy in the system.
A,U°
kJ
Transcribed Image Text:a In the reaction of two moles of gaseous sulfur dioxide and one mole of gaseous oxygen to form two moles of gaseous sulfur trioxide, two moles of products are formed from 3 moles of reactants. If this reaction is done at 1.00 atm pressure (and at 0 °C), the volume is reduced by 22.4 L. In this reaction, how much work is done on the system ( SO2, O2, SO3) by the surroundings? Work 2269.72 Correct Calculate the amount of work done on the surroundings using the equation w = -P × AV. J - = w = −101.3 J/(L · atm) × 1.00 atm × ( −22.4 L) E b The enthalpy change for this reaction is -197.9 kJ. Use this value, along with the answer to previous part, to calculate AU, the change in internal energy in the system. A,U° kJ
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