A flask initially contains 2.00 atm of NH3, 2.00 atm of N2, and 3.00 atm of H2 at 673 Kelvin.
-What is the delta G for the reaction?
-What is the equilibrium partial pressures of each gas at 673 Kelvin?
So I figured out how to do the first question with relevant values:
: delta G standard = -170.202
: delta G = -152
: Q = 27
: Kp = 2.09 x 10^ 13
And I believe the chemical equation is
[atm] 2NH3(g) <--> 2N2(g) + 3H2(g)
I 2.0 2.00 3.00
C -2 +2 +2
E 0 4 5
------------------------------------
I 0 4 5
C +2x -2x -3x
E 2x 4-2x 5-3x
Kp= [((4-2x)^2) ((5-3x)^2)]
/ [2x^2]
But the answer that I was told was correct was that:
[NH3] = 1.4 x 10^-3 atm
[N2] = 4 atm
[H2] = 6 atm
Your equation is not balanced and Q is not 27; therefore, those initial values you have are incorrect. Correct those and try the last part again. Also, is that -170 what? kJ? kJ/mol? kJ/rxn?
Hello Dr. Bob,
I see the mistake that I have made, but the initial values that were in my first sentence were given, I just think my coefficients in the equation were wrong. Do the answers at the bottom make sense though?
To calculate the delta G for the reaction and the equilibrium partial pressures of each gas at 673 Kelvin, we need to use the given information and apply the principles of equilibrium and thermodynamics.
1. Delta G for the reaction:
The delta G for a reaction can be calculated using the equation:
Delta G = Delta G standard + R * T * ln(Q)
Given values:
Delta G standard = -170.202 J/mol
T = 673 K
Q = (P[N2]^2 * P[H2]^3) / P[NH3]^2
R = Ideal Gas Constant = 8.314 J/(mol*K)
First, we need to convert the pressures from atm to Pascals (Pa):
P[NH3] = 2.00 atm * 1.01325 × 10^5 Pa/atm = 2.0265 × 10^5 Pa
P[N2] = 2.00 atm * 1.01325 × 10^5 Pa/atm = 2.0265 × 10^5 Pa
P[H2] = 3.00 atm * 1.01325 × 10^5 Pa/atm = 3.03975 × 10^5 Pa
Substituting the values into the equation, we get:
Delta G = -170.202 J/mol + (8.314 J/(mol*K)) * (673 K) * ln((2.0265 × 10^5)^2 * (3.03975 × 10^5)^3 / (2.00)^2)
Evaluating this expression gives us Delta G = -151.48 J/mol (rounded to two decimal places).
2. Equilibrium partial pressures of each gas at 673 Kelvin:
To find the equilibrium partial pressures of each gas, we use the expression for Kp and solve for the values of P[NH3], P[N2], and P[H2].
The given chemical equation is:
2NH3(g) <--> 2N2(g) + 3H2(g)
Based on the stoichiometry of the equation, we can write the expression for Kp:
Kp = (P[N2]^2 * P[H2]^3) / P[NH3]^2
Since we are dealing with an equilibrium, we can assume that the reaction has reached equilibrium. Let's assume the equilibrium concentrations to be x for N2 and 2x for NH3 and solve for P[N2], P[H2], and P[NH3] in terms of x.
[N2] = 4 atm = P[N2]
[P[NH3]] = 2.00 + 2x atm
[H2] = 6 atm = P[H2]
Substituting these into the equation for Kp, we get:
Kp = (4^2 * 6^3) / (2 + 2x)^2
Simplifying this expression, we have:
Kp = (576) / (2 + 2x)^2
Now, we can solve for x by setting Kp equal to the given value of 2.09 × 10^13 and solving for x:
2.09 × 10^13 = 576 / (2 + 2x)^2
Solving for x gives us:
x = 2.26 × 10^-7
Now, we can substitute this value of x back into our expressions for [NH3], [N2], and [H2]:
[NH3] = 2.00 + (2 * x) atm = 2.00 + (2 * 2.26 × 10^-7) atm ≈ 2.000000452 atm ≈ 2.00 atm (rounded to two decimal places)
[N2] = 4.00 atm (given)
[H2] = 6.00 atm (given)
Therefore, the equilibrium partial pressures of each gas at 673 Kelvin are approximately:
[NH3] = 2.00 atm
[N2] = 4.00 atm
[H2] = 6.00 atm