4. Consider the following reaction: A student performs a neutralization reaction at 298 K according to the following reaction:
NaHCO3 (8) + HCH, O2 (aq) → H20 (1) + CO2 (g) + NaC, H3O2 (aq)
The student adds 1.85 g of solid NaHCO3 into a flask containing 75.0 mL of 0.90 M HCH:02. The student
observes the formation of gas bubbles and also that the flask gets cooler to the touch as the reaction proceeds.
Considering all of the students observations that the student made, is the reaction thermodynamically favorable
at all temperatures, no temperatures, high temperatures or low temperatures. Justify (explain) your selection in
terms of AG

High temperatures

Explanation:

NaHCO₃ (8) + HCH,O₂ (aq) → H₂O (l) + CO₂ (g) + NaC,H₃O₂ (aq)

As the flask gets cooler to the touch as the reaction proceeds, the reaction is endothermic. This means that ΔH is positive (ΔH>0).

As a gas is formed (bubbles are formed), ΔS is positive (ΔS>0).

In terms of ΔG:

In order for the reaction to be thermodynamically favorable, ΔH has to be negative, thus:

The greater the temperature, the easier it would be for TΔS to be greater than ΔH.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.