Chemistry Help (1 Viewer)

Farhanthestudent005 · Sep 16, 2021

Will appreciate if I am assisted with this question. Thank you

jazz519 · Sep 16, 2021

deltaH =-ve as the reaction is exothermic and releases heat to the surroundings. Entropy is a measure of the disorder in a system. Here the reactants have more entropy because there are more particles than the products (2:1). This means entropy is decreasing, so deltaS = -ve

deltaG = deltaH - TdeltaS
deltaG = -ve - (+ve)(-ve)
deltaG = -ve +ve

As the temperature increases, the 2nd term (TdeltaS) increases in magnitude and so the deltaG is becoming more positive. As a result the spontaneity of the reaction is decreasing as the temperature goes up.

CM_Tutor · Sep 17, 2021

While I agree with @jazz519's reasoning, I think the conclusion drawn in an answer regarding spontaneity depends on perspective.

If you take the view that a reaction is spontaneous if $\Delta G < 0$ and non-spontaneous if $\Delta G > 0$ , then you might answer that:

as temperature increases, the value of $\Delta G$ will increase as $\Delta H < 0$ and $\Delta S < 0$
as $\Delta G = \Delta H - T\Delta S$ , there must be a critical temperature $T_{\text{crit}}$ at which $\Delta G = \Delta H - T_{\text{crit}}\Delta S = 0$
the reaction will be spontaneous so long as $T < T_{\text{crit}}$ , and will be non-spontaneous when $T > T_{\text{crit}}$

An example of this perspective is to consider the freezing of water, H₂O(l) ---> H₂O(s), which is exothermic ( $\Delta H < 0$ ) and entropically disfavoured ( $\Delta S < 0$ ), like the above example. For pure water under standard conditions, freezing is spontaneous at any temperature below 0 ^oC and non-spontaneous (i.e. the reverse process, melting, is spontaneous) at any temperature above 0 ^oC. Is it really meaningful to say that water freezes "more spontaneously" at (say) -5 ^oC than at -4 ^oC, or that spontaneity is greater?

Certainly, there are books / teachers / chemists who use language like "greater spontaneity" to mean that an already-negative $\Delta G$ is decreasing,. Describing this as a $\Delta G$ becoming "more negative" is linguistically debatable, as is the view of spontaneity as a relative concept that can be present in greater or lesser amounts, rather than as an absolute that is either present or absent under a given set of conditions. I favour the latter perspective and so would not say that " the deltaG is becoming more positive" when I mean that $\Delta G$ is negative but increasing towards zero as temperature increases because I don't see why -5 kJ mol^-1 is "more positive" than -10 kJ mol^-1.

I would answer:

As temperature increases, Gibbs Free Energy increases but the reaction remains spontaneous until the temperature passes the threshold at which $\Delta G$ becomes positive, where the reaction becomes non-spontaneous. We are told that the reaction is exothermic ( $\Delta H < 0$ ) and that the reaction proceeds (thus, $\Delta G < 0$ ), but the reaction is entropically disfavoured ( $\Delta S < 0$ ) as order increases when two species combine to form one in the same state. Since $\Delta G = \Delta H - T\Delta S$ , increasing temperatures will increase the value of the positive term ( $-T\Delta S$ ) and so cause $\Delta G$ to increase from its initial negative value, ultimately to become positive at some critical temperature, at which point the reaction ceases to be spontaneous.

Farhanthestudent005 · Sep 17, 2021

Thank you both of you! God bless you both

Chemistry Help (1 Viewer)

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