Calculate the work (in kJ) when 1.80 moles of methane react with excess oxygen at 389 K:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
I got 11661.26
Calculate the change in entropy (in J/K) when 70.1 g of nitrogen gas is heated at a constant pressure of 1.50 atm from 12.0 ºC to 63.0 ºC. (The molar specific heats are Cv is 20.8 J/(mol-K) and Cp is 29.1 J/(mol-K) .)
I got 63.2
Using the technique of the previous problem ΔE was found to be -2,000.00 kJ/mol of an unknown liquid hydrocarbon at 298 K. In another experiment it was determined that for each mole of hydrocarbon, 4 moles of oxygen gas are consumed and 9 moles of CO2 gas and 4 moles of H2O liquid are produced. Find ΔH per mole of this hydrocarbon (in kJ) at 298 K.
I did something along these lines:
Recall
ΔH = ΔE+PΔV = ΔE + ΔnRT for gases.
A comment on the last 2 problems. Normally, ΔH is calculated, then ΔE because it is easier to do experiments at constant pressure.
And got 10381.9
None of these are right unfortunately so please help! Thanks!
To calculate the work in the first problem, you need to use the equation:
Work = -ΔG
First, calculate the change in Gibbs free energy (ΔG) for the reaction using the equation:
ΔG = ΔH - TΔS
ΔH is the change in enthalpy and can be calculated using the enthalpies of formation for each compound involved in the reaction. The enthalpy of formation for CH4(g) is -74.8 kJ/mol, and for CO2(g) is -393.5 kJ/mol. The enthalpy change for the reaction is:
ΔH = (1 mol CO2 x -393.5 kJ/mol) - (1 mol CH4 x -74.8 kJ/mol) = -318.7 kJ
ΔS is the change in entropy, which can be calculated using the equation:
ΔS = ΣnS(products) - ΣnS(reactants)
The entropy values for each compound involved in the reaction can be found in a table. However, since only CO2 and CH4 have a significant contribution to ΔS, you can approximate it by using the entropy values for these two compounds. The entropy of CO2(g) is 213.7 J/(mol·K) and for CH4(g) is 186.3 J/(mol·K). The entropy change for the reaction is:
ΔS = (1 mol CO2 x 213.7 J/(mol·K)) - (1 mol CH4 x 186.3 J/(mol·K)) = 27.4 J/K
Now, substitute the values into the equation for ΔG:
ΔG = -318.7 kJ - (389 K x 27.4 J/K) = -328.5 kJ
Finally, calculate the work using the equation:
Work = -ΔG = 328.5 kJ
So, the correct value for the work is 328.5 kJ (not 11661.26 kJ).
For the second problem, you need to calculate the change in entropy (ΔS) using the equation:
ΔS = nCp·ln(T2/T1) - nR·ln(P2/P1)
where:
- n is the number of moles of the gas (which can be calculated using the given mass of nitrogen and its molar mass),
- Cp is the molar specific heat at constant pressure,
- R is the gas constant (8.314 J/(mol·K)),
- T2 and T1 are the final and initial temperatures in Kelvin,
- P2 and P1 are the final and initial pressures (since the pressure is constant, P2/P1 = 1).
First, calculate the number of moles of nitrogen gas (N2) using its molar mass:
molar mass of N2 = 28.02 g/mol
moles of N2 = 70.1 g / 28.02 g/mol = 2.498 mol
Now, substitute the values into the equation for ΔS:
ΔS = (2.498 mol · 29.1 J/(mol·K) · ln(336.15 K / 285.15 K)) - (2.498 mol · 8.314 J/(mol·K) · ln(1))
ΔS ≈ 2.498 mol · 29.1 J/(mol·K) · 0.163 - 0
ΔS ≈ 0.952 J/K
So, the correct value for the change in entropy is 0.952 J/K (not 63.2 J/K).
For the third problem, you need to use the equation:
ΔH = ΔE + PΔV = ΔE + ΔnRT
where:
- ΔH is the change in enthalpy,
- ΔE is the change in internal energy (given as -2,000.00 kJ/mol),
- Δn is the change in the number of moles of gas (which can be calculated from the stoichiometry of the reaction),
- R is the gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin.
From the balanced chemical equation, you can see that the change in the number of moles of gas is:
Δn = (9 mol CO2 + 4 mol H2O) - (4 mol O2 + 1 mol unknown hydrocarbon) = 4 mol CO2 + 4 mol H2O - 4 mol O2
Now, substitute the values into the equation for ΔH:
ΔH = -2,000.00 kJ/mol + (4 mol CO2 + 4 mol H2O - 4 mol O2) · 8.314 J/(mol·K) · 298 K
ΔH ≈ -2,000.00 kJ/mol + 498.432 kJ/mol
ΔH ≈ -1,501.57 kJ/mol
So, the correct value for ΔH per mole of the hydrocarbon is -1,501.57 kJ (not 10,381.9 kJ).