What is the standard entropy of reaction, the standard entropy of reaction, in J/K, for the combustion reaction of ethylene described in the previous questions? The temperature is 298.15K. (Enter your answer as a signed number in scientific notation without units.) My answers for the 2 previous questions are delta H = -1.41×10^3 kJ and delta G = -1.33×10^3 kJ. I have tried 3 times. The chemical equation is C2H4+3O2->2CO2+2H2O
What have you done those three times? Are you to calculate this from dG = dH - TdS or from standard S tables.
i am using dG=dH-tdS and i have tried
dG-dH/T,dG/dH+T,dG/dH-T. My answers are
-1.33,-1.0E3, -1.01E2
It appears to me that the answer is
-0.268 kJ. The problem wants the answer in J which is -268 and in scientific notation that is -2.68E2 (without the J unit).
To determine the standard entropy of reaction, we can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG is the standard Gibbs free energy change,
ΔH is the standard enthalpy change,
T is the temperature in Kelvin,
ΔS is the standard entropy change.
You have the values for ΔH and ΔG provided in your question. The temperature (T) is given as 298.15K.
First, let's convert the given values for ΔH and ΔG from kJ to J:
ΔH = -1.41 × 10^3 kJ = -1.41 × 10^3 × 10^3 J = -1.41 × 10^6 J
ΔG = -1.33 × 10^3 kJ = -1.33 × 10^3 × 10^3 J = -1.33 × 10^6 J
Now, rearrange the equation to solve for ΔS:
ΔS = (ΔH - ΔG) / T
Substituting the known values:
ΔS = (-1.41 × 10^6 J - (-1.33 × 10^6 J)) / 298.15 K
Now calculate the difference in the numerator:
ΔS = (-1.41 - (-1.33)) × 10^6 J / 298.15 K
Finally, divide the numerator by the temperature:
ΔS = -0.08 × 10^6 J / 298.15 K
Simplifying the scientific notation in the numerator:
ΔS = -8 × 10^4 J / 298.15 K
So, the standard entropy of reaction, ΔS, for the combustion reaction of ethylene is approximately -8 × 10^4 J/K (negative value indicates a decrease in entropy).