A chemical reaction with a negative enthalpy change and a negative entropy change ________. A) Must be nonspontaneous Is not pos
1 answer:
Answer:
Could be either spontaneous or nonspontaneous depending on the temperature
Explanation:
For a reaction to be spontaneous, Gibbs free energy must be negative.
The relation between enthalpy, entropy and temperature is as follows:
ΔG is change Gibbs free energy, ΔH is change in enthalpy, ΔS is change in entropy and T is temperature.
In the given condition, ΔH is negative and ΔS is negative. As ΔS is also negative then ΔG cannot be negative at all temperature (Particulary at high temperature) and thus, reaction could be nonspontaneous.
Therefore, in the given the reaction, could be either spontaneous or nonspontaneous depending on the temperature.
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This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:
First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:
Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:
This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
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