Skip to main content
Pearson+ LogoPearson+ Logo
Ch.19 - Chemical Thermodynamics
Back
Previous problem
Next problem
All textbooksBrown 14th EditionCh.19 - Chemical ThermodynamicsProblem 66

Chapter 19, Problem 66

A certain constant-pressure reaction is barely nonspontaneous at 45 °C. The entropy change for the reaction is 72 J/K. Estimate ΔH.

Verified Solution
Video duration:
2m
This video solution was recommended by our tutors as helpful for the problem above.
819
views
Was this helpful?

Video transcript

well everyone in this video, we're trying to determine the value of delta H. If the change of entropy is our delta S value for the reaction is 96 joules per kelvin. So of course we're gonna go ahead and relate these two with the gibbs free energy equation. But first things first, let's go ahead and convert the given temperature into our kelvin units. So it's given to us in Celsius 65°C, we can just go ahead and add 273.15 to give us 338.15 Kelvin's. So again the keyboard here is slightly non spontaneous. If we have that then let's go ahead and just draw this arrow here that the delta G is going to be greater than zero. So we have a positive value. So now bringing out the gibbs free energy equation. The delta G is equal to the change of entropy minus. The temperature, multiplied with the change of entropy. And of course as we said, that's going to be greater than zero. So our delta H. Is being subtracted by a temperature that we saw for two B 338.15 Calvin's then multiplied with our adult to us. And that's Given to us again, 96 jewels per Kelvin's, that's going to be greater than zero. So putting that into my calculator, it's going to be that delta H is subtracted from 3 to 462.4 jewels Again greater than zero. So it was kind of isolating our delta H. And we see here that we get the positive value of 3 to 462.4 jules. Let's go ahead and actually convert this into killer jewels. We can do a direct conversion for everyone. Killer jewels, we have 1000 jewels is here that the units of joules will cancel. So finalizing my delta age value to be greater than 32 killer whales. So if our delta S. Is positive and non spontaneous, the reaction must be an author Mick. And then we have this slightly non spontaneous reaction. The magnitude isn't greater than or can't be greater than R- 32 kg jewels. So to put this in writing, then the final answer is that the magnitude of delta H is not much greater then 32 killer jewels. And this right here is going to be my final answer for this problem. Thank you all so much for watching.
Previous problemNext problem
Related Practice
Textbook Question

Sulfur dioxide reacts with strontium oxide as follows: SO2(g) + SrO(g) → SrSO3(s) (a) Without using thermochemical data, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.

682
views
Textbook Question

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontaneous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T; (iv) spontaneous at high T but not spontaneous at low T. (c) N2F4(g) ⟶ 2  NF2(g) ΔH° = 85  kJ;  ΔS° = 198  J/K

839
views
Textbook Question

From the values given for ΔH° and ΔS°, calculate ΔG° for each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous? a. 2  PbS(s) + 3  O2(g) → 2  PbO(s) + 2  SO2(g) ΔH° = −844  kJ;  ΔS° = −165  J/K

2787
views
1
rank
1
comments
Textbook Question

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: CH3COOH(l) → CH3OH(g) + CO(g) By using data from Appendix C, calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that ΔH° and ΔS° do not vary with temperature.

806
views
Textbook Question

Consider the following reaction between oxides of nitrogen: NO2(g) + N2O(g) → 3 NO(g) (a) Use data in Appendix C to predict how ΔG for the reaction varies with increasing temperature.

516
views
Textbook Question

Consider the following reaction between oxides of nitrogen: NO2(g) + N2O(g) → 3 NO(g) (b) Calculate ΔG at 800 K, assuming that ΔH° and ΔS° do not change with temperature. Under standard conditions is the reaction spontaneous at 800 K?

654
views