Calcium carbonate's decomposition, non-spontaneous at 298 K and spontaneous above about 1110 K, gets argued from enthalpy and entropy across the SC190 Unit 8 thermodynamics application. Searches like "sc 190 unit 8 assignment example", "sc190 unit 8 sample" and "sc190 unit 8 example" land here.
What a finished SC190 Unit 8 thermodynamics application looks like
Three to four pages built on one reaction, CaCO3 forming CaO and CO2, with a data table of standard enthalpies of formation and standard entropies at the top. The enthalpy calculation sums products minus reactants: -635.1 plus -393.5, minus -1206.9, gives +178.3 kJ. The entropy calculation does the same with 39.7, 213.7 and 92.9 J/mol K, giving +160.5 J/K, positive because a gas forms from a solid. With entropy converted to kilojoules, the free energy at 298.15 K is 178.3 minus 298.15 times 0.1605, or +130.4 kJ, so the reaction does not proceed on its own. Setting the free energy to zero gives a crossover near 1111 K, about 838 degrees Celsius, and a check at 1200 K returns -14.3 kJ. A short section turns the 298 K value into an equilibrium constant of about 1.4 x 10^-23.
How a SC190 Unit 8 example is structured
The application states its question first, whether this process runs spontaneously and under what conditions. Enthalpy and entropy are each calculated separately, with signs interpreted in a sentence before they are combined. The combination step is where most care goes: units are reconciled explicitly, since tables give entropy in joules and enthalpy in kilojoules, and the temperature is in kelvins. A sign table follows, the four combinations of enthalpy and entropy signs with what each means for temperature dependence, and the reaction is placed in its row. The crossover temperature is computed and then tested at a temperature above it. An assumptions paragraph acknowledges that enthalpy and entropy are treated as constant with temperature, which makes the crossover an estimate. The equilibrium constant section ties the result back to the term's earlier equilibrium work.
Products minus reactants, twice
Enthalpy and entropy are each summed from the table with coefficients applied, giving +178.3 kJ and +160.5 J/K, and each sign is explained before the two are combined.
Joules converted before combining
Entropy enters the Gibbs equation as 0.1605 kJ/K, and the application writes the conversion out, since a missing factor of 1000 makes this endothermic reaction look spontaneous.
Placed in the sign table
Positive enthalpy with positive entropy puts the reaction in the row that becomes spontaneous at high temperature, which predicts the answer before any crossover is calculated.
A crossover, then a test above it
Dividing 178.3 kJ by 0.1605 kJ/K gives about 1111 K, and evaluating the free energy at 1200 K returns a negative value that confirms the prediction.
Free energy tied to K
The relationship between standard free energy and the natural log of K turns +130.4 kJ into K of about 1.4 x 10^-23, a number that explains why limestone stays limestone.
Where marks go in SC190 Unit 8
Units are where most points go: entropy in joules subtracted from enthalpy in kilojoules, which turns +130.4 kJ into a large negative number and reverses the conclusion. Temperatures in Celsius inside the Gibbs equation come next. Coefficients ignored when summing formation values, or elements in their standard states given nonzero enthalpies, cost accuracy at the first step. Spontaneity confused with speed is a steady deduction, since a negative free energy says a process can proceed, not that it will happen quickly. Crossover temperatures reported as exact, with no mention of the constant-enthalpy assumption, draw a comment in many rubrics. Answers that call an endothermic reaction impossible, rather than non-spontaneous at the stated temperature, lose the reasoning credit the unit is built around, and a final free energy with no units closes the list.
Get a SC190 Unit 8 example written to your instructions
Share the reaction or process your SC190 Unit 8 prompt asks about, together with its data table and rubric. The application calculates enthalpy, entropy and free energy with units reconciled, places the process in the sign table and finds any crossover temperature. Allow 24-48h for delivery; a first application is free.
SC190 Unit 8 questions, answered
Does a negative free energy mean the reaction happens quickly?
No. Free energy tells you whether a process can proceed on its own, not how fast. Diamond converting to graphite has a negative free energy at room conditions and effectively never happens on a human timescale. Speed is a kinetics question, which is why the rate work earlier in SC190 and this unit answer different questions.
Why is the crossover temperature only an estimate?
Because the calculation assumes enthalpy and entropy stay constant as temperature rises, using their 298 K values at 1100 K and above. Both change somewhat with temperature, so the true crossover differs from the calculated one. For a course application the assumption is standard, but stating it shows the grader you know the result is approximate.
Where do the standard values come from?
From the appendix of the course text or a standard thermodynamic table, and the application should cite whichever it uses. Values differ slightly between sources, by a few tenths of a kilojoule or a joule per kelvin, which shifts the crossover by a few kelvins. Using the table your section provides keeps your answer consistent with the grader's key.