Calculate the free energy change for this reaction at 25 °C. Is the reaction spontaneous? (Assume that all reactants and products are in their standard states.) C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) ΔH°_rxn= -2217 kJ; ΔS°_rxn = 101.1 J/K

Question

Accepted Answer

hey everyone today we're being asked to consider the following reaction and are given the entropy of the reaction as well as its entropy, assuming that all the reactions and the products are in their standard state. We are asked to calculate the delta G and determine if the reaction is spontaneous or not. Now let's write out the terms that we have and the formula we need to find delta G. Using the entropy and the entropy of a reaction. We can use the formula delta G. Is equal to delta H. The reaction minus T. Delta S. Where H. Is the entropy and S. Is the entropy and T. Is the temperature in kelvin. So let's go ahead and write out our terms. We know that the delta H. Of the reaction Is -479.4 killer jewels. And the delta S. The entropy Is negative 189.2 jewels per kelvin. Now keep in mind that we need to express delta G. The free energy of the reaction in kill jewels. So we need to convert this to kill a jewels. Which is very simple Because remember for every one jewel we have 10 to the three or sorry for everyone killer jewel, we have 10 to the third jewels. So we can multiply it by that factor we one kill a jewel For every 10 to the third jewels gives us zero, sorry negative 0.1892 killer jewels per mole. And and I'll use green for this. Our temperature temperature At standard state is 25°C. But we need this in Kelvin. So we add to 73.153.15, which is equal to 200, 298.15 Kelvin. So plugging this all back into our equation, we have delta G. Of the reaction is equal to -4 79.4 Killer Jewels minus 298.15 Kelvin times -0.1892 killer jewels per per kelvin. Let me fix that up here as well. This is kilo joules per kelvin. And simplifying this, we get the delta g of the reaction is negative 423 zero Killer Jewels. And because Delta G is less than zero, it is spontaneous spawn titanius. So I hope this helps. And I look forward to seeing you in the next one.

Calculate the free energy change for this reaction at 25 °C. Is the reaction spontaneous? (Assume that all reactants and products are in their standard states.) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH°rxn = -2217 kJ; ΔS°rxn = 101.1 J/K

Verified Solution

Calculate the free energy change for this reaction at 25 °C. Is the reaction spontaneous? (Assume that all reactants and products are in their standard states.) C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) ΔH°_rxn= -2217 kJ; ΔS°_rxn = 101.1 J/K