example 3 - 8\nfind the difference between molar δh and δu for the combustion of octane at 298 k.\nc₈h₁₈(l)…

example 3 - 8\nfind the difference between molar δh and δu for the combustion of octane at 298 k.\nc₈h₁₈(l) + 25/2 o₂(g) → 8 co₂(g) + 9 h₂o(l)\nδh_reaction ≅ δu_reaction + rtδn_gases (at constant t)\nδh - δu = rtδn_gases

example 3 - 8\nfind the difference between molar δh and δu for the combustion of octane at 298 k.\nc₈h₁₈(l) + 25/2 o₂(g) → 8 co₂(g) + 9 h₂o(l)\nδh_reaction ≅ δu_reaction + rtδn_gases (at constant t)\nδh - δu = rtδn_gases

Answer

Explanation:

Step1: Identify the formula

The formula for the relationship between $\Delta H$ and $\Delta U$ at constant temperature is $\Delta H-\Delta U = RT\Delta n_{gases}$.

Step2: Calculate $\Delta n_{gases}$

For the reaction $\text{C}8\text{H}{18}(l)+\frac{25}{2}\text{O}_2(g)\rightarrow8\text{CO}_2(g) + 9\text{H}2\text{O}(l)$, the number of moles of gaseous products minus the number of moles of gaseous reactants is $\Delta n{gases}=8-\frac{25}{2}=\frac{16 - 25}{2}=- 4.5$ mol.

Step3: Substitute values

We know that $R = 8.314\ J/(mol\cdot K)$ and $T = 298\ K$. Substitute $\Delta n_{gases}=-4.5$ mol, $R = 8.314\ J/(mol\cdot K)$ and $T = 298\ K$ into the formula $\Delta H-\Delta U=RT\Delta n_{gases}$. $\Delta H-\Delta U=8.314\ J/(mol\cdot K)\times298\ K\times(-4.5\ mol)$ $\Delta H-\Delta U=8.314\times298\times(- 4.5)\ J$ $\Delta H-\Delta U=-11187.51\ J=-11.19\ kJ$ (rounded to two - decimal places)

Answer:

$-11.19\ kJ$