Calculate the delta G of the reaction using the following information.
4HNO3+ 5N2H4 --> 7N2+ 12H2O
Delta H(Kj/mol) -133.9 50.6 0 -285.8
Delta S(J/mol x K) 266.9 121.2 191.6 70.0
Well, calculating delta G requires both delta H and delta S. But since you've kindly provided them, let's get to it.
First, we need to convert delta H from KJ/mol to J/mol. So, delta H = -133.9 × 1000 = -133900 J/mol.
Now, let's calculate delta S. The reaction has 4 moles of HNO3 and 5 moles of N2H4, so the change in moles of gas is: Δn = (7+12) - (4+5) = 10 moles.
Therefore, delta S = (10 mol) × (70.0 J/mol·K) = 700 J/mol·K.
Now, we can calculate delta G using the equation: ΔG = ΔH - TΔS. Since you didn't provide the temperature (T), I'm just going to use room temperature, approximately 298 K.
ΔG = -133900 J/mol - (298 K × 700 J/mol·K) = -51500 J/mol.
However, I must clarify that my calculations might contain errors because, you know, I'm a clown bot. So please double-check everything just to be safe.
To calculate the delta G (Gibbs free energy) of the reaction, you can use the equation:
ΔG = ΔH - TΔS
where ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.
Given:
ΔH(HNO3) = -133.9 Kj/mol
ΔH(N2H4) = 50.6 Kj/mol
ΔH(N2) = 0 Kj/mol
ΔH(H2O) = -285.8 Kj/mol
ΔS(HNO3) = 266.9 J/mol x K
ΔS(N2H4) = 121.2 J/mol x K
ΔS(N2) = 191.6 J/mol x K
ΔS(H2O) = 70.0 J/mol x K
First, calculate the total ΔH and ΔS for the reaction by summing the products and subtracting the reactants:
ΔH(total) = (7 * ΔH(N2)) + (12 * ΔH(H2O)) - (4 * ΔH(HNO3)) - (5 * ΔH(N2H4))
= (7 * 0) + (12 * -285.8) - (4 * -133.9) - (5 * 50.6)
= -2000.6 Kj/mol
ΔS(total) = (7 * ΔS(N2)) + (12 * ΔS(H2O)) - (4 * ΔS(HNO3)) - (5 * ΔS(N2H4))
= (7 * 191.6) + (12 * 70.0) - (4 * 266.9) - (5 * 121.2)
= 617.2 J/mol x K
Assuming the reaction is at standard temperature, T = 298 K.
Now substitute the values into the ΔG equation:
ΔG = ΔH - TΔS
= -2000.6 - (298 * 617.2)
= -2000.6 - 183914.7
= -185915.3 Kj/mol
Therefore, the ΔG of the reaction is -185915.3 Kj/mol.
To calculate the delta G (change in Gibbs free energy) of the reaction, you can use the equation:
Delta G = Delta H - T * Delta S
where Delta H is the change in enthalpy, T is the temperature in Kelvin, and Delta S is the change in entropy.
Given:
Delta H (delta enthalpy):
-133.9 kJ/mol for 4HNO3
50.6 kJ/mol for 5N2H4
0 kJ/mol for 7N2
-285.8 kJ/mol for 12H2O
Delta S (delta entropy):
266.9 J/mol x K for 4HNO3
121.2 J/mol x K for 5N2H4
191.6 J/mol x K for 7N2
70.0 J/mol x K for 12H2O
We need to solve it at a specific temperature, so let's assume T = 298.15 K.
Now, let's calculate the delta G using the given equation:
Delta G = [(-133.9 kJ/mol * 4) + (50.6 kJ/mol * 5)] - [(266.9 J/mol x K * 4) + (121.2 J/mol x K * 5)]
+ [0 kJ/mol * 7] - (191.6 J/mol x K * 7) - (285.8 kJ/mol * 12) - (70.0 J/mol x K * 12)
First, convert the units to match:
Delta G = [(-133.9 kJ/mol * 4) + (50.6 kJ/mol * 5)] - [(266.9 J/mol x K * 4) + (121.2 J/mol x K * 5)]
+ [0 kJ/mol * 7] - (191.6 J/mol x K * 7) - (285.8 kJ/mol * 12) - (70.0 J/mol x K * 12)
Next, solve for the change in Gibbs free energy:
Delta G = [-535.6 kJ + 253.0 kJ] - [1067.6 J + 606.0 J]
+ [0 kJ] - [1341.2 J] - [-3429.6 kJ] - [840 J]
Delta G = -282.6 kJ + 746.4 J - 1341.2 J - 3429.6 kJ - 840 J
Delta G = -4276.2 kJ - 595 J
Finally, convert the units back to kilojoules:
Delta G ≈ -4.2762 x 10^6 J - 0.595 kJ
Delta G ≈ -4.2762 x 10^6 J - 0.000595 x 10^6 J
Delta G ≈ -4.2768 x 10^6 J
Therefore, the delta G of the reaction is approximately -4.2768 x 10^6 J, or -4.2768 MJ.
Use the information to calculate dG for each reactant and each product with dG = dH - TdS, then
dGrxn = (n*dG products) - (n*dG reactants)
Post your work if you get stuck.