Question

What is the difference between $\Delta H$ and $\Delta U$ at $298\, K$ for the following reaction?
$ C _2 H _4(g)+3 O _2(g) \rightarrow 2 CO _2(g)+2 H _2 O ( I )$
$\Delta H=-1410.0\, kJ$

• A. $-14.8\, k J$
• B. $-2.45 \,k J$
• C. $-4.95\, k J$
• D. $-7.30\, kJ$

Answer 1

Answer: (C)

In the given reaction,
$ C _2 H _4(g)+3 O _2(g) \rightarrow 2 CO _2(g)+2 H _2 O (l)$
$ \Delta H=-1410.0\, kJ$
$ \text { For ideal gas, } \Delta H=\Delta U+R T\left(\Delta n_g\right)$
$\text { where, } R=8.314 J / \text { molK } $
$ T=298 \, K$
$\Delta n_g=\text { number of gaseous products - number of gaseous reactants }$
$ =2-4=-2$
$ \therefore-1410=\Delta U+[8.314 \times 298(-2)] $
$ \Delta U=-1405.045 \, kJ $
$ \sim e q-1405.05 \, kJ$
$ \text { Now, } \Delta H-\Delta U=-1410-(-1405.05)=-4.95\, kJ$