If the reaction Fe2N(s) + 3/2H2(g) -> 2Fe(s) +NH3(g) comes to equilibrium at a total pressure of 1 bar, analysis of the gas shows that at 700. and 800.K, PNH3/PH2=2.165 and 1.083 respectively, if only H2(g) was initially present in the gas phase and Fe2N(s) was in excess.
a) calculate Kp at 700 and 800K.
b)Calculate DeltaSr at 700 and 800K and DeltaHr assuming that it is independent of temperature.
c. Calculate deltaGr for this reaction at 298.15K
a) calculate Kp at 700 and 800K.
b)Calculate DeltaSr at 700 and 800K and DeltaHr assuming that it is independent of temperature.
c. Calculate deltaGr for this reaction at 298.15K
el procedimiento ?
To solve the given questions, we first need to understand the equation:
Fe2N(s) + 3/2H2(g) ⇌ 2Fe(s) + NH3(g)
a) To calculate Kp at 700K and 800K, we can use the given information. The ratio of PNH3 to PH2 is equal to the coefficient ratio of NH3 to H2 in the balanced equation.
At 700K:
PNH3/PH2 = 2.165
This means that the concentration of NH3 is 2.165 times greater than that of H2.
At 800K:
PNH3/PH2 = 1.083
This means that the concentration of NH3 is 1.083 times greater than that of H2.
To calculate Kp, we use the ideal gas law:
Kp = (PNH3)^2 / (PH2)^(3/2)
At 700K:
Kp = (2.165)^2 / (1)^^(3/2)
At 800K:
Kp = (1.083)^2 / (1)^(3/2)
b) To calculate ΔSr (change in entropy) at 700K and 800K, we can use the equation:
ΔSr = ΔS°products - ΔS°reactants
Since we have excess Fe2N(s) and only H2(g) initially present, the entropy change is only due to the formation of NH3(g):
ΔSr = R * ln(PNH3 / P°)
Where R is the gas constant and P° is the standard pressure.
At 700K:
ΔSr = R * ln(2.165 / P°)
At 800K:
ΔSr = R * ln(1.083 / P°)
Since the question assumes ΔHr (change in enthalpy) is independent of temperature, we do not need to calculate it.
c) To calculate ΔGr (change in Gibbs free energy) at 298.15K, we can use the equation:
ΔGr = ΔG° + RT * ln(Q)
Where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature, and Q is the reaction quotient.
At equilibrium, ΔGr = 0, so we can set ΔGr = 0 and solve for ΔG°:
0 = ΔG° + RT * ln(Kp)
ΔG° = - RT * ln(Kp)
Finally, we can substitute the value of ΔG° into the equation to calculate ΔGr at 298.15K.
a) To calculate Kp at 700K and 800K, we can use the equation:
Kp = (PNH3^2) / (PH2^(3/2))
Given that PNH3/PH2 is 2.165 and 1.083 at 700K and 800K respectively, we can substitute these values into the equation to calculate Kp.
At 700K:
Kp = (2.165^2) / (1)^1.5
Kp = 4.679
At 800K:
Kp = (1.083^2) / (1)^1.5
Kp = 1.179
b) To calculate DeltaSr (change in entropy) at 700K and 800K, we need the formula:
DeltaSr = -R * ln(Kp)
Where R is the gas constant (8.314 J/(mol·K)). Given that DeltaHr (change in enthalpy) is assumed to be independent of temperature, we don't need to calculate it.
At 700K:
DeltaSr = -8.314 * ln(4.679)
DeltaSr = - 8.314 * 1.539
DeltaSr = -12.804 J/(mol·K)
At 800K:
DeltaSr = -8.314 * ln(1.179)
DeltaSr = -8.314 * 0.166
DeltaSr = -1.381 J/(mol·K)
c) To calculate deltaGr (change in Gibbs free energy) at 298.15K, we can use the equation:
deltaGr = deltaHr - T * DeltaSr
Given that deltaHr is assumed to be independent of temperature, we just need to calculate DeltaSr.
At 298.15K:
DeltaSr = -12.804 J/(mol·K) (from part b)
deltaGr = deltaHr - (298.15K * -12.804 J/(mol·K))
(Note: deltaGr is assumed to be at standard conditions, so T = 298.15K)
This calculation requires the value of deltaHr to get the final answer. If the value of deltaHr is given, you can substitute it into the equation to calculate deltaGr at 298.15K.