How do i calculate:
What is is ΔG° at 25 °C?
2O3(g) → 3 O2(g)
ΔH° = -284 kJ ΔS° = 139 J/K
i know the equation is:
ΔG°= ΔH°- t (ΔS)
im not getting right answer
nvm i got it had to conver delta S J/k to kj/k
To calculate ΔG° at 25 °C using the equation ΔG° = ΔH° - T(ΔS°), you need to make sure that the temperature (T) is in Kelvin (K) rather than Celsius (°C). So, to convert 25 °C to Kelvin, add 273 to it: 25 + 273 = 298 K.
Now you have all the values you need:
ΔH° = -284 kJ
ΔS° = 139 J/K
T = 298 K
To make the units consistent, convert the ΔH° value to J:
ΔH° = -284 kJ × 1000 J/1 kJ = -284,000 J
Now you can calculate ΔG°:
ΔG° = ΔH° - T(ΔS°)
ΔG° = -284,000 J - 298 K × 139 J/K = -284,000 J - 41,462 J = -325,462 J
The ΔG° at 25 °C is -325,462 J.
To calculate ΔG° at 25 °C, you are correct to use the equation:
ΔG° = ΔH° - T(ΔS°)
Where:
ΔG° = Standard Gibbs Free Energy Change
ΔH° = Standard Enthalpy Change
T = Temperature in Kelvin (25 °C = 25 + 273.15 K)
ΔS° = Standard Entropy Change
Let's substitute the given values into the equation:
ΔH° = -284 kJ (Convert to J by multiplying by 1000)
= -284,000 J
ΔS° = 139 J/K
T = 25 °C + 273.15 K
= 298.15 K
Now we can calculate ΔG°:
ΔG° = (-284,000 J) - (298.15 K)(139 J/K)
= -284,000 J - 41,453.85 J
= -325,453.85 J
Therefore, ΔG° at 25 °C is approximately -325,453.85 J.