For each pair of compounds, predict which species will have the higher molar entropy at 298 K and 1 bar. Be sure to state your reasons for your choice.
1. N2(g) vs. F2(g)
2. ICl (g) vs. BrF(g)
3. 6Li(s) vs. 7Li(s
To predict which species will have the higher molar entropy at 298 K and 1 bar, we need to consider the factors that influence entropy. Here are the reasons for each choice:
1. N2(g) vs. F2(g)
To determine the molar entropy, we need to know the standard molar entropy value for each compound. In this case, N2(g) is diatomic nitrogen, and F2(g) is diatomic fluorine. The standard molar entropy tends to increase with the number of atoms in a compound. Since N2(g) has a higher number of atoms than F2(g), we can predict that N2(g) will have a higher molar entropy at 298 K and 1 bar.
2. ICl(g) vs. BrF(g)
Similar to the previous case, we need to compare the standard molar entropy values for these compounds. ICl(g) is a diatomic molecule, whereas BrF(g) is a compound consisting of one bromine atom and one fluorine atom. The number of atoms in a compound again influences the entropy. Hence, BrF(g) will have a higher molar entropy compared to ICl(g) at 298 K and 1 bar.
3. 6Li(s) vs. 7Li(s)
In this case, we are comparing solid elemental lithium isotopes. Both 6Li(s) and 7Li(s) are atoms in their solid-state, and the molar entropy of an element in its standard state is always zero. Therefore, both 6Li(s) and 7Li(s) will have the same molar entropy of zero at 298 K and 1 bar.
Remember, these predictions are based on the comparison of standard molar entropy values and the principle that an increase in the number of atoms in a compound generally leads to a higher entropy.