Question

Calculate the enthalpy change when 2.38gcarbon monoxide$\left(CO\right)$vaporizes at its normal boiling point. Use data from Table 12.2.

Short answer

The enthalpy change, $\Delta$H=512.8J.

Step-by-step solution

Given data

The mass of oxygen, m(O)=2.38g .

Concept of thermochemistry

The study of heat energy related to chemical reactions and/or phase changes such as melting and boiling is known as thermochemistry.

Calculate the number of moles

The number of moles can be determined as:

It is given that, M(CO)=28.01g/mol.

It will be determined by the formula, n=m/M. where, n is the number of moles, m is mass, and M is molar mass.

Now, put the value of mass and molar mass carbon mono oxide in the above formula.

N(CO)=2.38

28.01

= 0.849mol

Calculation of enthalpy change (ΔH)

The difference between the change in enthalpy and the standardized value, in this case, the enthalpy change is determined with the help of a formula,$\Delta H=n\times \Delta H_m.$$\Delta H=n\times \Delta H_m.$ .

Put the value of the given data in the above expression.

$\begin{array}{l}\Delta H=0.0849\times 6.04\times {10}^3\\ \Delta H=512.8J\end{array}$

Therefore, the enthalpy change, $\Delta H=512.8J$.