What is the ∆g for the following reaction between o3 and o2? 3o2(g) 2o3(g) given: o3: ∆h = 285.4 kj ∆s = -137.14 j/k)109.5 kj208.3 kj300.2 kj326.3 kj

We’ll be using the equation:
dG = dH - TdS (replace ‘d’ with triangle)

I’m going to assume 0 degrees Celsius.
At 0 C (273 K):

dG = dH - TdS
dG = (285,400 J) - (273 K)(-137.14 J/K)
dG = 285,400 J + 37,439.2 J
dG = 322,839.2 J or 322.84 kJ

The dG of this reaction is +322.84 kJ. This reaction is not considered spontaneous.

This answer, in this instance, would be D. If the temperature used in the question is not 0 degrees C, replace the temperature that I used for calculation with the Kelvin temperature given in the problem (K = C + 273), and simplify to find the answer.

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.