Question

The molar heat capacity is represented in the temp range 298 to 400 k by the emprical expression 14.73+0.1272 calculate the enthalpy if formation of ethane at 350 k from its valu at 298 k

Answer 1

Answer:

$\Delta _f H=-81.1\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the enthalpy of formation of ethane at 298 K is about -84 kJ/mol, via the Kirkoff's law, we can compute it a 350 K as shown below:

$\Delta _f H=\Delta _f H\° +\int\limits^{350K}_{298K} {14.73+0.1272T} \, dx$

It means we can integrate to get:

$\Delta _f H=\Delta _f H\° +14.73\frac{J}{mol*K} (350K-298K)+\frac{0.1272}{2}\frac{J}{mol*K^2} (350K^2-298K^2)$

Now, we can plug in the enthalpy of formation in consistent units to obtain:

$\Delta _f H=-84,000\frac{J}{mol} +14.73\frac{J}{mol*K} (350K-298K)+\frac{0.1272}{2}\frac{J}{mol*K^2} (350K^2-298K^2)\\\\\Delta _f H=-81,091\frac{J}{mol}\\\\\Delta _f H=-81.1\frac{kJ}{mol}$

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