What is the value of the equilibrium constant, K, for the given reaction? Express the answer in terms of log(K) 4XY+3Z2 * 2Z3X2 +2Y2 Substance ΔΗΡΟ/mol) sº (mol-K) 139.0 73.0 Y2 - 82.0 37.0 Za = 25.0 78.0 Z₃ X₂ 245.0 137.0 =1135 kmol 5.9 kymol.
Answers
Answer:
The equilibrium constant expression for a gas reaction is
K
c
=
[NO]
4
[H
2
O]
6
[NH
3
]
4
[O
2
]
5
.
The balanced chemical equation corresponding to this expression is
4NO+6H
2
O⇌4NH
3
+5O
2
.
Answer:
Explanation:
To calculate the equilibrium constant, we need to use the formula:
K = e^(-ΔG°/RT)
where ΔG° is the standard free energy change, R is the gas constant, and T is the temperature in Kelvin.
First, we need to calculate ΔG° for the reaction. We can use the formula:
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
where ΔG°f is the standard free energy of formation, and n is the stoichiometric coefficient.
For the reactants:
ΔG°f(4XY) = 0 (by definition)
ΔG°f(3Z2) = 3*(-82.0 kJ/mol) = -246.0 kJ/mol
For the products:
ΔG°f(2Z3X2) = 2*(245.0 kJ/mol) = 490.0 kJ/mol
ΔG°f(2Y2) = 2*(-37.0 kJ/mol) = -74.0 kJ/mol
Therefore,
ΔG° = (2245.0 - 237.0) - 3*(-82.0) = 750.0 kJ/mol
Next, we need to convert ΔG° to J/mol:
ΔG° = 750.0 * 1000 = 750000 J/mol
The temperature is not given, so we will assume it to be 298 K.
Now we can calculate K using the formula:
K = e^(-ΔG°/RT) = e^(-750000/(8.314*298)) = 2.53 x 10^(-18)
Finally, we can express log(K) as:
log(K) = -18.60
Therefore, log(K) for the given reaction is -18.60.