The equilibrium constant Kp is 5.00×10^17 at 25◦C for the reaction C2H4(g) + H2(g)<->C2H6(g). From this information, calculate ∆G0 at 25◦C
dGo = -RT*lnK
R is 8.314 and T is 298.15. You know K. Solve for dGo.
Oh, calculating ∆G0, huh? Well, let me balance the equation between my comedic side and scientific knowledge. So, C2H4 and H2 walked into a bar, and things got heated. Eventually, they formed C2H6, leaving all the other molecules green with envy. Now, let's calculate the ΔG0, the energy that fueled this chemical comedy.
To calculate ∆G0, we can use the equation ∆G0 = -RTln(Kp), where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25°C = 298 K), and ln represents the natural logarithm.
∆G0 = - (8.314 J/(mol·K)) × 298 K × ln(5.00×10^17)
Now, I'll let you do the math part because I don't want to take all the glory from you. But remember, even if the calculation gets too intense, don't forget to smile and keep your chemistry humor in balance!
To calculate ∆G° at 25°C using the equilibrium constant (Kp) for the reaction C2H4(g) + H2(g) ⇌ C2H6(g), we'll use the relationship:
∆G° = -RT ln(Kp)
where:
- ∆G° is the change in Gibbs free energy under standard conditions,
- R is the gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin, and
- Kp is the equilibrium constant.
First, convert 25°C to Kelvin:
T = 25 + 273.15 = 298.15 K
∆G° = - (8.314 J/(mol·K)) * ln(Kp)
Now substitute the equilibrium constant (Kp) into the equation and solve:
∆G° = - (8.314 J/(mol·K)) * ln(5.00 × 10^17)
Using a calculator or software, calculate the natural logarithm (ln) of 5.00 × 10^17:
ln(5.00 × 10^17) = 39.7
∆G° = - (8.314 J/(mol·K)) * 39.7
= - 329.8 J/mol
Therefore, the value of ∆G° at 25°C is approximately -329.8 J/mol.
To calculate ∆G0 using the equilibrium constant (Kp), you can use the formula:
∆G0 = -RT ln(Kp)
Where:
- ∆G0 is the standard Gibbs free energy change
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (25 + 273.15 = 298.15 K)
- ln(Kp) is the natural logarithm of the equilibrium constant
Step 1: Calculate ln(Kp)
ln(Kp) = ln(5.00×10^17)
Step 2: Calculate ∆G0
∆G0 = - (8.314 J/mol·K) x (298.15 K) x ln(5.00×10^17)
By substituting the values into the equation and solving, you can find the value of ∆G0.