Consider the rxn during the Haber Process:
N2 + 3H2 ---> 2NH3
The equilibrium constant is 3.9 x 10^(5) at 300 K and 1.2 x 10^(-1) at 500 K.
Find Delta H reaction and Delta S for reaction
Won't the van't Hoff equation allow you to calculate dHo?
Then use K to calculate dGo = -RTlnK, and use that for dGo = dH-TdS to calculate dS.
The solubility of AgCl(s) in water at 25 C is 1.33x10^-5 M and its deltaH of solution is 65.7 kJ/mol .
What is its solubility at 60.1 C ?
Please explain how to solve for this!!
To find ΔH (enthalpy change) and ΔS (entropy change) for the reaction, we can use the Van't Hoff equation, which relates the change in equilibrium constant with temperature to the enthalpy and entropy changes.
The Van't Hoff equation is given by:
ln(K2/K1) = ΔH/R * (1/T1 - 1/T2)
Where:
K1 is the equilibrium constant at temperature T1
K2 is the equilibrium constant at temperature T2
ΔH is the enthalpy change
R is the gas constant (8.314 J/mol·K)
T1 and T2 are the initial and final temperatures (in Kelvin)
Let's calculate ΔH and ΔS step by step for the given reaction.
Step 1: Calculate ΔH using the Van't Hoff equation
We have K1 = 3.9 x 10^5 and T1 = 300 K
K2 = 1.2 x 10^(-1) and T2 = 500 K
ln(K2/K1) = ΔH/R * (1/T1 - 1/T2)
ln(1.2 x 10^(-1)/3.9 x 10^5) = ΔH/8.314 * (1/300 - 1/500)
Solving the equation above for ΔH, we can find:
ΔH = -26.7 kJ/mol
So, the enthalpy change for the reaction is -26.7 kJ/mol.
Step 2: Calculate ΔS using the equation:
ΔS = ΔH/T - Rln(K)
We already know ΔH = -26.7 kJ/mol, T = 300 K (since we're calculating at the initial temperature), and R = 8.314 J/mol·K.
To find ΔS at the initial temperature, plugging in the values into the equation, we get:
ΔS = (-26.7 kJ/mol) / (300 K) - (8.314 J/mol·K) * ln(3.9 x 10^5)
Simplifying the equation above, we can find:
ΔS = -89.0 J/mol·K
So, the entropy change for the reaction at the initial temperature is -89.0 J/mol·K.
To find the values of ΔH (enthalpy change) and ΔS (entropy change) for the reaction, we can use the Van't Hoff equation:
ln(K2/K1) = -(ΔH/R) * (1/T2 - 1/T1)
where:
K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively,
ΔH is the enthalpy change of the reaction,
R is the gas constant (8.314 J/(mol·K)),
T1 and T2 are the initial and final temperatures, respectively.
Let's substitute the given values into the equation and solve for ΔH:
For T1 = 300 K and K1 = 3.9 x 10^(5):
ln(K2/(3.9 x 10^(5))) = -(ΔH/8.314) * (1/500 - 1/300)
To solve for ΔH, we need to determine ln(K2/(3.9 x 10^(5))). Rearranging the equation, we get:
ln(K2/(3.9 x 10^(5))) = (ΔH/8.314) * (1/300 - 1/500)
Now, substitute the given values for T2 = 500 K and K2 = 1.2 x 10^(-1):
(ΔH/8.314) * (1/300 - 1/500) = ln((1.2 x 10^(-1))/(3.9 x 10^(5)))
Rearranging and solving for ΔH, we have:
ΔH = [ln((1.2 x 10^(-1))/(3.9 x 10^(5)))] / [(1/300 - 1/500)] * 8.314
Now, let's calculate the value of ΔH using the formula and the given values.
Once you have calculated ΔH, you can find ΔS using the equation:
ΔS = ΔH / T
where ΔS is the entropy change, ΔH is the enthalpy change, and T is the temperature in Kelvin.