Methanol (CH3OH) can be made by controlled oxidation of methane.
CH4(g) + 1/2 O2(g) �¨ CH3OH(g)
I found that the
ƒ¢H�‹ is -126.4kJ/K and the ƒ¢S�‹ is -51.2, and the ƒ¢G�‹ at 298 K is 111kJ
Yet how do you
Calculate the temperature at equilibrium? please show steps if possible
It will be delta H over delta S because at equilibrium delta G is equal to 0
To calculate the temperature at equilibrium, you can use the equation:
ΔG = ΔH - TΔS
Since the temperature at equilibrium is not specified, we'll use T to represent it. Rearranging the equation, we have:
T = (ΔH - ΔG)/ΔS
Substituting the given values, we have:
T = (-126.4 kJ/K - 111 kJ)/(51.2 J/K) * (kJ/1000 J)
Converting kJ to J:
T = (-126.4 kJ/K - 111 kJ)/(51.2 J/K) * (1 kJ/1000 J)
T = (-126.4 kJ - 111 kJ)/(51.2 J) * (1 kJ/1000 J)
T = -237.4 kJ / 51.2 J * (1 kJ/1000 J)
T = -4.64 K
The result is -4.64 K. However, negative temperatures are not physically meaningful in this context. Therefore, it is important to note that the equilibrium temperature cannot be determined using the given values.
To calculate the temperature at equilibrium, you can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
At equilibrium, the value of ΔG will be zero. So the equation becomes:
0 = ΔH - TΔS
Rearranging the equation to solve for T:
T = ΔH / ΔS
Now, let's substitute the given values:
ΔH = -126.4 kJ/mol
ΔS = -51.2 J/(mol·K)
First, we need to convert the units of ΔH to J/mol:
ΔH = -126.4 kJ/mol * 1000 J/kJ = -126,400 J/mol
Now, substitute the values into the equation:
T = (-126,400 J/mol) / (-51.2 J/(mol·K))
T = 2470 K
Therefore, the temperature at equilibrium is 2470 Kelvin.