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shutvik [7]
3 years ago
14

If δh°rxn and δs°rxn are both positive values, what drives the spontaneous (favored) reaction and in what direction at standard

conditions?
Chemistry
1 answer:
Arada [10]3 years ago
8 0
In thermodynamics and physical chemistry, the Gibb's free energy is the criterion for spontaneity. If the Gibb's free energy, denoted as ΔG, is negative, then the reaction is spontaneous. If it is positive, it is non-spontaneous. To estimate ΔG, there is a derived relationship between Gibb's free energy, enthalpy and entropy. The equation is

ΔG=ΔH-TΔS, where ΔH and ΔS is the enthalpy and entropy of the reaction, and T is the temperature at which the reaction happens.
So if ΔH and ΔS are both positive, in order to make ΔG negative, T must be very high. In this way, TΔS would be larger than ΔH.

Therefore, a reaction with a positive ΔH and ΔS will be spontaneous at high temperatures.
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alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

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