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

`\Delta _f H=-81.1(kJ)/(mol)`

Step-by-step 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(J)/(mol*K) (350K-298K)+(0.1272)/(2)(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(J)/(mol) +14.73(J)/(mol*K) (350K-298K)+(0.1272)/(2)(J)/(mol*K^2) (350K^2-298K^2)\\\\\Delta _f H=-81,091(J)/(mol)\\\\\Delta _f H=-81.1(kJ)/(mol)`

Best regards!