A certain constant-pressure reaction is nonspontaneous at 37°C. The entropy change for the reaction is 83 J/K. What can you conclude about the sign of ÄH?
What can you conclude about the magnitude of ÄH?
To determine the sign of ΔH (enthalpy change) in this situation, we can use the relationship between entropy change (ΔS) and ΔH. The relationship is given by the equation:
ΔG = ΔH - TΔS
where ΔG is the Gibbs free energy change and T is the temperature in Kelvin.
Since the reaction is nonspontaneous at 37°C, the Gibbs free energy change (ΔG) must be positive. In other words, the reaction is not favored to proceed in the forward direction.
Considering this, if we rearrange the equation above to solve for ΔH:
ΔH = ΔG + TΔS
We know that ΔS (entropy change) is given as 83 J/K. Since the reaction is nonspontaneous, ΔG must be positive. As a result, to ensure that ΔH is also positive and maintain a positive ΔG, the calculated value of TΔS should be greater than ΔG.
From the information given, we know that ΔS is positive (83 J/K), so TΔS will be positive as well. Since ΔG must be positive to keep the reaction nonspontaneous, we can infer that the sum of ΔG + TΔS is also positive. Therefore, we can conclude that ΔH is positive (or endothermic) in this particular reaction.
Regarding the magnitude of ΔH, we cannot determine its exact value based on the provided information. The given entropy change (ΔS) of 83 J/K does not provide information about the specific magnitude of ΔH. It only tells us the sign of ΔH, which is positive (endothermic). The magnitude of ΔH could be smaller or larger depending on the precise values of ΔG and T.