Question

Calculate the temperature at which the following reaction becomes feasible:4HCl + O2 -> 2Cl2 + 2H2O Enthalpy: +280 kJ/molEntropy: +235 J/KmolA. 515.0 KB. 45.0 KC. 1191.49 KD. 65.80 K

Answer 1

C. 1191.49 K.

Step-by-step explanation:

Entropy and Free Energy => Temperature and Free Energy.

We can use the following formula that involves enthalpy (ΔH), entropy (ΔS), and temperature (T):

`T=(\Delta H)/(\Delta S).`

You can see that the units of enthalpy are kJ/mol, and entropy is J/Kmol. You can realize that we cannot do the calculation with these units because the energy unit is not equal (kJ≠J), so we can convert the enthalpy, 280 kJ/mol to J/mol remembering that 1 kJ equals 1000 J like this:

`280(kJ)/(mol)\cdot\frac{1000\text{ J}}{1\text{ kJ}}=280000(J)/(mol).`

So now we have ΔH = 280000 J/mol and ΔS = 235 J/Kmol and we can replace these data in the given formula:

`\begin{gathered} T=\frac{280000\text{ }(J)/(mol)}{235(J)/(Kmol)}, \\ \\ T=1191.489\text{ K}\approx1191.49\text{ K.} \end{gathered}`

The answer would be that the temperature is C. 1191.49 K.