Question

estimate the difference between $delta h_{reaction}$ and $delta u_{reaction}$ for the combustion of liquid butanol ($c_{4}h_{9}oh$) in excess $o_{2}$, at $t = 298 k$. take into consideration that water produced in this reaction is in the liquid state. j/mol

Explanation:

Step1: Write the balanced combustion equation

$C_4H_9OH(l)+6O_2(g)
ightarrow4CO_2(g)+5H_2O(l)$

Step2: Determine $\Delta n_{gas}$

$\Delta n_{gas}=n_{products,gas}-n_{reactants,gas}=4 - 6=- 2$

Step3: Use the relationship $\Delta H=\Delta U+\Delta n_{gas}RT$

The difference $\Delta H-\Delta U=\Delta n_{gas}RT$. Given $R = 8.314\ J/(mol\cdot K)$ and $T = 298\ K$.
$\Delta H-\Delta U=-2\times8.314\ J/(mol\cdot K)\times298\ K$
$\Delta H-\Delta U=-4955.144\ J/mol\approx - 4955\ J/mol$

Answer:

$-4955$