Use the Table of Thermodynamic Data to decide which of the following reactions have
K>1 at 298 K:
(i) HCl(g) + NH3(g)-->NH4Cl(s)
(ii) 2 Al2O3(s) + 3 Si(s)-->3 SiO2(s) + 4 Al(s)
(iii) Fe(s) + H2S(g)-->FeS(s) + H2(g).
Which of the products of these reactions are favored by a rise in temperature?
Chris, Terry, and the others. It really helps us help you if you use the same screen name each session.
I don't know what data you have available in your tables but I would try
dGorxn = (n*dGoproducts - n*dGoreactants) and if dGrxn < 0, then K>1.
For the last part, do the same for dHorxn = (n*dHoproducts) - (n*dHoreactants) and if dHorxn is - the reaction is exothermic.
To determine which of the given reactions have K>1 at 298 K, we need to look up the standard Gibbs free energy change (ΔG°) for each reaction and calculate the equilibrium constant (K) using the equation:
ΔG° = -RT ln(K)
where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.
The reactions and their respective standard Gibbs free energy changes at 298 K are:
(i) HCl(g) + NH3(g) --> NH4Cl(s)
ΔG° = -177.78 kJ/mol
(ii) 2 Al2O3(s) + 3 Si(s) --> 3 SiO2(s) + 4 Al(s)
ΔG° = -2113.2 kJ/mol
(iii) Fe(s) + H2S(g) --> FeS(s) + H2(g)
ΔG° = -84.12 kJ/mol
Using the equation ΔG° = -RT ln(K), we can rearrange it to solve for K:
K = e^(-ΔG° / RT)
Substituting the values of ΔG° and the given temperature (298 K), we can calculate the equilibrium constants (K) for each reaction:
(i) K = e^(-(-177.78 kJ/mol) / (8.314 J/(mol·K) * 298 K))
= e^(67290.872 / 2472.872)
≈ 1.1 x 10^24
(ii) K = e^(-(-2113.2 kJ/mol) / (8.314 J/(mol·K) * 298 K))
= e^(802420.08 / 2472.872)
≈ 2.7 x 10^349
(iii) K = e^(-(-84.12 kJ/mol) / (8.314 J/(mol·K) * 298 K))
= e^(31945.128 / 2472.872)
≈ 3.4 x 10^13
Based on the calculated equilibrium constants (K), all of the given reactions have K>1 at 298 K.
Now, to determine which products are favored by a rise in temperature, we can look at the ΔG° values of the reactions.
If ΔG° is negative, it means the reaction is exothermic and the products are favored by a decrease in temperature. Conversely, if ΔG° is positive, the reaction is endothermic and the products are favored by an increase in temperature.
(i) ΔG° = -177.78 kJ/mol (negative), so the formation of NH4Cl is favored by a decrease in temperature.
(ii) ΔG° = -2113.2 kJ/mol (negative), so the formation of SiO2 and Al is favored by a decrease in temperature.
(iii) ΔG° = -84.12 kJ/mol (negative), so the formation of FeS and H2 is favored by a decrease in temperature.
In summary:
(i) The products NH4Cl(s) is favored by a decrease in temperature.
(ii) The products SiO2(s) and Al(s) are favored by a decrease in temperature.
(iii) The products FeS(s) and H2(g) are favored by a decrease in temperature.
To determine which of the reactions have K>1 at 298 K, we need to make use of the Table of Thermodynamic Data. This table provides the standard Gibbs free energy change (ΔG°) for various chemical reactions at a specific temperature, usually 298 K.
For each reaction, we can calculate the standard Gibbs free energy change (ΔG°) using the equation:
ΔG° = ΣΔG°(products) - ΣΔG°(reactants)
If ΣΔG° is negative, it means the reaction is favored in the forward direction (K>1). If ΣΔG° is positive, it means the reaction is favored in the reverse direction (K<1).
Now, let's calculate the ΔG° for each reaction:
(i) HCl(g) + NH3(g) --> NH4Cl(s)
ΔG° = ΔG°(NH4Cl) - ΔG°(HCl) - ΔG°(NH3)
(ii) 2 Al2O3(s) + 3 Si(s) --> 3 SiO2(s) + 4 Al(l)
ΔG° = ΔG°(3 SiO2) + ΔG°(4 Al) - ΔG°(2 Al2O3) - ΔG°(3 Si)
(iii) Fe(s) + H2S(g) --> FeS(s) + H2(g)
ΔG° = ΔG°(FeS) + ΔG°(H2) - ΔG°(Fe) - ΔG°(H2S)
Next, we can consult the Table of Thermodynamic Data, which provides the ΔG° for each compound involved in the above reactions at 298 K. By substituting these values into the respective equations and performing the calculations, we can determine whether ΔG° is positive or negative for each reaction. If ΔG° is negative, it means K>1 at 298 K.
To determine which products are favored by a rise in temperature, we need to consider the effect of temperature on the equilibrium constant. The general rule is that an increase in temperature favors an endothermic reaction (generally, reactions with positive ΔH). For exothermic reactions (generally, reactions with negative ΔH), an increase in temperature favors the formation of reactants.
By examining the enthalpy change (ΔH) for each reaction, which can also be found in the Table of Thermodynamic Data, we can determine the effect of temperature on equilibrium. If ΔH is positive, a rise in temperature will favor the formation of products. If ΔH is negative, a rise in temperature will favor the formation of reactants.
By following these steps of calculating ΔG° and considering the effect of temperature on ΔH, we can determine the reactions with K>1 at 298 K and which products are favored by a rise in temperature.