A certain constant-pressure reaction is nonspontaneous at 49 ∘C. The entropy change for the reaction is 71 J/K .
What can you conclude about the magnitude of ΔH?
answers :ΔH>23kJ
ΔH>220kJ
ΔH<71J/K
ΔH>71J/K
ΔH<23kJ
ΔH<220kJ
Hmm, let me think for a moment... *taps imaginary clown nose*
Ah, I've got it! Based on the given information, we know that the reaction is nonspontaneous at 49 ∘C. The entropy change for the reaction is 71 J/K. So, if we're talking about the magnitude of ΔH (that's the change in enthalpy), we can safely conclude that ΔH must be greater than 71 J/K.
Now I know you're probably thinking, "Clown Bot, give me a specific answer!" Well, I'm sorry to disappoint you, but I can't provide an exact value based on what we know so far. So, the closest answer to our conclusion is ΔH > 71 J/K. No worries, this clown's got your back with some entertaining answers! 🎈
Based on the given information, we know that the reaction is nonspontaneous at 49 °C and the entropy change (∆S) is 71 J/K. In order for a reaction to be spontaneous at constant pressure, both the entropy change and the enthalpy change (∆H) must be favorable.
Since the reaction is nonspontaneous, we can conclude that ∆H is not favorable (negative) at 49 °C. Therefore, we can eliminate the choices ΔH > 23 kJ, ΔH > 220 kJ, and ΔH > 71 J/K because they suggest a favorable (positive) ∆H.
This leaves us with the choices ΔH < 71 J/K, ΔH < 23 kJ, and ΔH < 220 kJ.
However, since we do not have any information about the specific magnitudes of ∆H, we cannot definitively conclude whether ∆H is smaller or larger than 71 J/K, 23 kJ, or 220 kJ.
Therefore, the correct answer is: ΔH < 71 J/K, ΔH < 23 kJ, and ΔH < 220 kJ.
To determine the relationship between the entropy change (ΔS) and the enthalpy change (ΔH) for a reaction, we can use the following equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
ΔS is the change in entropy
T is the temperature in Kelvin
Since the reaction is nonspontaneous at 49 ∘C, it implies that ΔG is positive. In other words:
ΔG > 0
Now, substituting the values into the equation:
0 = ΔH - TΔS
Since we are given that ΔS = 71 J/K and T = 49 ∘C (which needs to be converted to Kelvin by adding 273), we can rewrite the equation as:
0 = ΔH - (49 + 273) * 71
Simplifying:
0 = ΔH - 22248
Rearranging the equation to solve for ΔH:
ΔH = 22248
Therefore, we can conclude that the magnitude of ΔH is greater than 22.2 kJ (since 22248 J = 22.248 kJ).
Therefore, the correct answer is ΔH > 22 kJ.
dG = dH - TdS
To be not spontaneous dG must be -
dH = ?
T is 49+273 = 322
- = dH - 322(71)
- = dH - 22862 J
- = dH - 22,862.
What is the value for dH so that a negative value is obtained when added to 22.3 kJ.