2 H2S (g) + SO2 (g) <===> ƒnƒn3 S (s) + 2 H2O (g)
1) At 298 K, the standard enthalpy change, H¢X, for the reaction represented below is -145 kilojoules.
(a) Predict the sign of the standard entropy change, S¢X, for the reaction. Explain the basis foryour prediction.
(b) At 298 K, the forward reaction (i.e., toward the right) is spontaneous. What change, if any, would occur in the value of G¢X for this reaction as the temperature is increased? Explain your reasoning using thermodynamic principles.
(c) What change, if any, would occur in the value of the equilibrium constant, Keq, for the situation described in (b)? Explain your reasoning.
(d) The absolute temperature at which the forward reaction becomes non spontaneous can be predicted. Write the equation that is used to make the prediction. Why does this equation predictonly an approximate value for the temperature?
(a) Since the standard enthalpy change, H¢X, is negative, it means that the reactants have a higher enthalpy than the products. This suggests that the reaction releases energy and becomes more stable. Entropy, on the other hand, is a measure of disorder or randomness in a system. Generally, when a reaction becomes more stable, it tends to have a decrease in entropy because the products are more orderly than the reactants. Therefore, based on the given information, we can predict that the sign of the standard entropy change, S¢X, for the reaction is negative.
(b) According to thermodynamic principles, the spontaneity of a reaction at a given temperature can be determined by the change in Gibbs free energy, G¢X. The equation relating the change in Gibbs free energy, temperature, and the change in enthalpy is:
ΔG = ΔH - TΔS
Since the forward reaction is spontaneous at 298 K, it means that ΔG is negative. When the temperature is increased, the ΔS term (change in entropy) has a negative sign due to part (a) and the equation becomes:
ΔG = ΔH - TΔS
If ΔH and ΔS remain constant, an increase in temperature (T) would make the negative (-TΔS) term larger. As a result, ΔG would become less negative or even positive, indicating that the reaction would become less spontaneous or even non-spontaneous.
(c) The equilibrium constant, Keq, is related to the change in Gibbs free energy by the equation:
ΔG = -RT*ln(Keq)
where R is the gas constant and T is the temperature.
As we saw in part (b), an increase in temperature would cause the value of ΔG to become less negative or even positive. Since ΔG is in the denominator of the equation, a smaller (less negative) ΔG would make the right side of the equation (the natural logarithm term) larger. This means that the value of Keq would decrease. Therefore, an increase in temperature would cause a decrease in the value of the equilibrium constant, Keq.
(d) The absolute temperature at which the forward reaction becomes non-spontaneous can be predicted using the equation:
ΔG = ΔH - TΔS
To find the temperature, we can set ΔG equal to zero. This gives us:
0 = ΔH - TΔS
Rearranging the equation, we get:
T = ΔH/ΔS
This equation can only provide an approximate value for the temperature because it assumes that ΔH and ΔS remain constant over the entire temperature range, which might not always be true. Additionally, this equation does not account for other factors such as changes in heat capacity with temperature. Nevertheless, it gives us a rough estimate of the temperature at which the forward reaction becomes non-spontaneous.
(a) The sign of the standard entropy change, ∆S°X, for the reaction can be predicted based on the stoichiometry of the reaction. When one mole of reactants is converted into three moles of products, the number of gaseous molecules increases. This increase in the number of gaseous molecules usually leads to an increase in entropy. Therefore, we can predict that the standard entropy change, ∆S°X, for the reaction is likely to be positive.
(b) At 298 K, if the forward reaction is spontaneous (i.e., ∆G°X < 0), increasing the temperature would cause the value of ∆G°X to become more negative. This is because an increase in temperature favors the endothermic direction of the reaction, which is the forward direction in this case. According to the equation ∆G°X = ∆H°X - T∆S°X, an increase in temperature (T) would decrease the positive term T∆S°X, making ∆G°X more negative.
(c) The equilibrium constant, Keq, is related to the values of ∆G°X and temperature (T) by the equation ∆G°X = -RTln(Keq), where R is the ideal gas constant. If we increase the temperature, according to the previous explanation, ∆G°X becomes more negative. Since ∆G°X is in the exponent in the equation, a more negative ∆G°X leads to a larger Keq value. Therefore, increasing the temperature would cause the value of the equilibrium constant, Keq, to increase.
(d) The equation used to predict the absolute temperature at which the forward reaction becomes non-spontaneous is:
∆G°X = ∆H°X - T∆S°X = 0
Rearranging the equation gives:
T = (∆H°X)/(∆S°X)
This equation predicts only an approximate value for the temperature because it assumes that ∆H°X and ∆S°X are constant over the temperature range of interest, which might not be the case. In reality, both enthalpy and entropy can vary with temperature, so the prediction is an approximation.
(a) To predict the sign of the standard entropy change (ΔS°X) for the given reaction, we need to consider the number of moles of reactants and products. In this case, there are two moles of reactants (2 H2S and SO2) and three moles of products (3S and 2 H2O). Since the number of moles of products is greater than the number of moles of reactants, we can infer that there is an increase in the randomness or disorder of the system during the reaction. This suggests that the standard entropy change (ΔS°X) for the reaction will be positive.
(b) Spontaneity of a reaction is determined by the change in Gibbs free energy (ΔG). At constant temperature, the relationship between ΔG, ΔH, and ΔS is given by the equation:
ΔG = ΔH - TΔS
Where T is the temperature in Kelvin. Since the standard enthalpy change (ΔH°X) for the reaction is negative (-145 kJ), we can conclude that ΔH < 0. For a spontaneous reaction, ΔG must be negative. As temperature increases, the value of -TΔS will become more significant in the equation. In order for ΔG to remain negative, the magnitude of ΔS must increase as well. Therefore, as the temperature increases, the value of ΔG°X for this reaction will become more negative.
(c) The equilibrium constant (K_eq) for a reaction is related to the Gibbs free energy change using the equation:
ΔG° = -RT ln(K_eq)
Where R is the gas constant and T is the temperature in Kelvin. As we discussed earlier, as the temperature increases, the value of ΔG° becomes more negative. Since ΔG° is directly related to ln(K_eq), a decrease in ΔG° will result in an increase in K_eq. Therefore, as the temperature increases, the value of the equilibrium constant (K_eq) for the reaction will increase.
(d) The absolute temperature (T) at which the forward reaction becomes non-spontaneous can be predicted using the equation:
ΔG = ΔH - TΔS
When a reaction becomes non-spontaneous, ΔG becomes positive. So, we can rewrite the equation as:
T = ΔH / ΔS
This equation allows us to calculate an approximate value for the temperature at which the forward reaction becomes non-spontaneous. However, it's important to note that this equation only predicts an approximate value because it does not take into account factors such as pressure or the concentration of reactants and products, which can affect the spontaneity of a reaction.
a) because the moles of gas decreases, there is less disorder and the delta s value is negative.
b) the delta g value would become less negative and eventually positive with enough heat.
c) the keq would become lower.
d) (delta g)= (delta h) - T(delta s)
the delta g value would be replaced with 0.