The change in entropy for a reaction is 3.0 J/(mol·K) at 25°C. Calculate the change in free energy for the reaction when ΔH = –394 kJ/mol and determine whether it will occur spontaneously at this temperature.

Question

The change in entropy for a reaction is 3.0 J/(mol·K) at 25°C. Calculate the change in free energy for the reaction when ΔH = –394 kJ/mol and determine whether it will occur spontaneously at this temperature.

Is the answer -395 kJ/mol?

and second question: using this equation below to calculate the change in enthalpy for this reaction: C8H18(l) + 25/2O2(g) --> 8C02(g) + 9H2O(g)

H2(g) + 1/2O2(g) --> H2O(g) DeltaH = -241.8 kJ
C(s) + O2(g) --> CO2(g) DeltaH = -393.5 kJ
8C(s) +9H2(g) --> C8H18(l) DeltaH = -224.13 kJ

Is the answer -5100 kJ? Sorry for asking a lot I'm not good at these questions :(

Answer 1

1.

dG = dH - TdS
dG = -394 kJ - 298(0.003) = -394.89 kJ = -395 kJ.

2. Yes, I obtained -5100.2 kJ which rounds to -5100 kJ.

Answer 2

To calculate the change in free energy for a reaction, you can use the equation ΔG = ΔH - TΔS, where ΔG is the change in free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

For the first question, the change in entropy (ΔS) is given as 3.0 J/(mol·K). The change in enthalpy (ΔH) is -394 kJ/mol. The temperature (T) is 25°C, which needs to be converted to Kelvin by adding 273:

T = 25°C + 273 = 298 K

Substituting these values into the equation, we get:

ΔG = -394 kJ/mol - (298 K)(3.0 J/(mol·K))

To convert 3.0 J/(mol·K) to kJ/(mol·K), divide it by 1000:

ΔG = -394 kJ/mol - (298 K)(0.003 kJ/(mol·K))

Calculating:

ΔG = -394 kJ/mol - 0.894 kJ/mol

ΔG ≈ -394.894 kJ/mol

So, the change in free energy for the reaction is approximately -394.894 kJ/mol.

To determine whether the reaction will occur spontaneously at this temperature, we can look at the sign of ΔG. For a reaction to be spontaneous at constant temperature and pressure, ΔG must be negative. In this case, -394.894 kJ/mol is indeed negative, so the reaction should occur spontaneously at 25°C.

For the second question, we need to use the provided enthalpy values to calculate the change in enthalpy for the reaction C8H18(l) + 25/2O2(g) -> 8CO2(g) + 9H2O(g).

Taking each compound in the reaction equation and using the given enthalpy values, we get:

8C(s) + 9H2(g) -> C8H18(l) ΔH = -(-224.13 kJ) = 224.13 kJ (note the negative sign cancels out)

25/2O2(g) -> 25/2CO2(g) ΔH = - (25/2)(-393.5 kJ) = 4918.75 kJ

8C(s) + 16H2(g) -> 8CO2(g) + 8H2O(g) ΔH = -8(-393.5 kJ) = 3148 kJ

Adding up the enthalpy values gives:

ΔH = 224.13 kJ + 4918.75 kJ + 3148 kJ = 8289.88 kJ

So, the change in enthalpy for the reaction C8H18(l) + 25/2O2(g) -> 8CO2(g) + 9H2O(g) is approximately 8289.88 kJ.

Please note that the answer is not -5100 kJ. If you have any further questions, feel free to ask!