The following reaction is reactant-favored at equilibrium at room temperature.
COCl2(g) = CO(g) + Cl2(g)
Will raising or lowering the temperature make it product-favored?
We need to know if the reaction is endothermic or exothermic. We can calculate dH for the reaction through the heats of formation as follows:
dHorxn = (n*dHo products) - (n*dHo reactants)
Look in your text/notes and find the dHo vaues for CO, COCl2 and Cl2. I've done that below.
n = 1; COCl2 = -223 kJ/mol
n = 1; CO = -110.5 kJ/mol
n = 1; Cl2 = 0
Substitute into the equation above as follows:
dHorxn = (-110.5 + 0) - (-223)
dHorxn = -110.5+223 = + ?
So we know the reaction is endothermic and we can rewrite the equation as
COCl2 + heat ==> CO + Cl2.
If we raise T it shifts the equilibrium to the right. Whether it is shifted enough to make it product favored depends upon delta G. To calculate dG,
Do the same thing I've done above and calculate dSo rxn. With that information you can calculate dG.
dG = dH - TdS.
Set dG = 0, plug in dH and dS and calculate T at which the reaction is spontaneous to the right. I hope this makes sense to you. You may just be looking for raising T will make it product favored. My rationale is that raising T will shift the reaction to the right; however, whether that makes it FAVORED is true in my mind only if it makes the reaction spontaneous.
To determine whether raising or lowering the temperature will make the reaction product-favored, we need to analyze the reaction's enthalpy change (ΔH).
If the reaction is exothermic (ΔH < 0), meaning it releases heat, increasing the temperature will shift the reaction towards the reactants, making it reactant-favored.
On the other hand, if the reaction is endothermic (ΔH > 0), meaning it absorbs heat, increasing the temperature will favor the products, making it product-favored.
Unfortunately, without information about the enthalpy change for the given reaction, we cannot definitively say whether raising or lowering the temperature will make it product-favored.
To determine if raising or lowering the temperature will make the reaction product-favored, we need to consider the sign of the enthalpy change (∆H) for the reaction.
If ∆H is negative, the reaction is exothermic, meaning it releases heat. In this case, lowering the temperature will make the reaction product-favored because it will shift the equilibrium towards the products to counteract the decrease in temperature.
If ∆H is positive, the reaction is endothermic, meaning it absorbs heat. In this case, raising the temperature will make the reaction product-favored because it will shift the equilibrium towards the products to counteract the increase in temperature.
To determine the sign of ∆H for the reaction, we can use the concept of bond energies. The breaking of bonds requires energy, and the formation of bonds releases energy. We need to compare the energy required to break the bonds in the reactants (COCl2) to the energy released when the new bonds form in the products (CO and Cl2).
By looking up the bond energies of the relevant bonds involved in the reaction, we can calculate the total energy required to break the bonds in the reactants and the total energy released when the new bonds form in the products.
Based on the comparison of these energies, we can determine whether ∆H is positive or negative, and thus whether raising or lowering the temperature will make the reaction product-favored.
However, without specific bond energies and temperature information, we cannot definitively determine if raising or lowering the temperature will make the reaction product-favored in this case.