When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 169.5 kJ are
absorbed and PΔV for the vaporization process is equal to 14.5 kJ.
What is the change of internal energy?
What is the change of enthalpy?
Is the change in entropy positive or negative for this process?
If the pressure would be only ½ atm, what are the changes of internal energy and enthalpy during vaporization (approximately)?
How much energy is absorbed?
Write the definition of the Gibbs Free Energy.
At the lower pressure, will the boiling point be higher or lower.
What is the change of internal energy?
What is the change of enthalpy?
Is the change in entropy positive or negative for this process?
If the pressure would be only ½ atm, what are the changes of internal energy and enthalpy during vaporization (approximately)?
How much energy is absorbed?
Write the definition of the Gibbs Free Energy.
At the lower pressure, will the boiling point be higher or lower.
1 answer:
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Answer:
1.346 v
Explanation:
1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:
(oxidation) →
E°=0.337 v
(reduction) →
E°=1.679 v
(overall) +8H^{+}_{(aq)}→
E°=1.342 v
2) Nernst Equation
Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:
Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.
The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give
E=1.346