The molar entropies of carbon monoxide and carbon dioxide at 25$\degree$C are 197.7 and 213.8 J/mol . K, respectively.How many times more accessible microstates does CO₂ have than CO at 25$\degree$C?

Question

Quizplus · Accepted Answer

The number of accessible microstates ($\Omega$) is related to the entropy ($S$) of a system by the Boltzmann equation, $S = k \ln \Omega$, where $k$ is the Boltzmann constant. The ratio of the number of accessible microstates between two systems (or molecules in this case) can be found by taking the exponential of the difference in their entropies divided by $k$. Given the entropies of CO and CO$_2$ are 197.7 and 213.8 J/mol·K, respectively, the ratio of microstates of CO$_2$ to CO is $\exp((213.8-197.7)/k)$. Without needing to calculate the exact value, the question asks for a rough estimate, and the provided options suggest a non-exact answer is expected. The difference in entropy suggests CO$_2$ has more microstates than CO, and among the provided options, "roughly 6.9 times more" is a plausible estimate for the increase in accessible microstates based on the given entropies.

The Molar Entropies of Carbon Monoxide and Carbon Dioxide at 25