Energy is priceless

Chemistry Level 2

A chemical reaction occurs at room temperature, which is about 293 K 293 \text{ K} (kelvin), What can we say about the change in enthalpy, Δ H , \Delta H, and the change in entropy, Δ S , \Delta S, of this reaction, assuming that the change in entropy is positive?

Δ H > Δ S \Delta H > \Delta S Δ H < Δ S \Delta H < \Delta S Δ H Δ S > 293 K \dfrac{\Delta H}{\Delta S} > 293 \text{ K} Δ H Δ S < 293 K \dfrac{\Delta H}{\Delta S} < 293 \text{ K} Δ H Δ S = 293 K \dfrac{\Delta H}{\Delta S} = 293 \text{ K} Δ H = Δ S \Delta H = \Delta S None of these choices

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1 solution

Steven Yuan
Aug 16, 2017

The Gibbs free energy Δ G \Delta G of a reaction determines whether a reaction will occur at a given temperature. It can be calculated with the formula,

Δ G = Δ H T Δ S , \Delta G = \Delta H - T \Delta S,

where T T is the temperature in kelvin. When the Gibbs free energy of the reaction is negative, the reaction is said to be exergonic , which basically means that the reaction occurs in the direction it is written in.

Since the given reaction occurs at room temperature, we know that Δ G < 0 \Delta G < 0 and T = 293 K . T = 293 \text{ K}. Thus,

Δ H ( 293 K ) Δ S < 0 Δ H < ( 293 K ) Δ S Δ H Δ S < 293 K . \begin{aligned} \Delta H - (293 \text{ K}) \Delta S &< 0 \\ \Delta H &< (293 \text{ K}) \Delta S \\ \dfrac{\Delta H}{\Delta S} &< 293 \text{ K}. \end{aligned}

The direction of the less-than sign does not change because we are given that Δ S > 0. \Delta S > 0. We conclude that Δ H Δ S < 293 K \boxed{\dfrac{\Delta H}{\Delta S} < 293 \text{ K}} is the only conclusion we can derive from the given information.

(EDIT: I noticed that the answer choices comparing Δ H \Delta H and Δ S \Delta S directly are not appropriate because they have different units and, thus, cannot be compared directly. Can this be fixed somehow?)

"exorgonic" is termed as spontaneos in India :P

Md Zuhair - 3 years, 8 months ago

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