Question

What is the standard enthalpy of the reaction $ CO + {H_2}O \to C{O_2} + {H_2} $ ?

Answer 1

Hint: The measure of energy in a thermodynamic process is enthalpy. The overall content of heat in a system is enthalpy, which is proportional to the system's internal energy along with the product of volume and pressure.

Complete answer:
We need to look up the standard enthalpy values for each material in the reaction to solve for the standard enthalpy.
The standard enthalpy value of $ CO $ is $ - 110.5kJ{\left( {mol} \right)^{ - 1}} $
The standard enthalpy value of $ {H_2}O $ is $ - 241.826kJ{\left( {mol} \right)^{ - 1}} $
The standard enthalpy value of $ C{O_2} $ is $ - 393.5kJ{\left( {mol} \right)^{ - 1}} $
The standard enthalpy value of $ {H_2} $ is $ 0kJ{\left( {mol} \right)^{ - 1}} $
Once we have obtained these numbers, put them into the standard enthalpy formation equation and solve for the standard enthalpy.
 $ \Delta {H_{reaction}} = \Delta {H_{products}} - \Delta {H_{reac\tan ts}} $
To find the standard enthalpy value of the products, we have to add the standard enthalpy value of $ C{O_2} $ and the standard enthalpy value of $ {H_2} $
 $ \Delta {H_{products}} = \Delta {H_{C{O_2}}} + \Delta {H_{{H_2}}} $
To find the standard enthalpy value of the reactants, we have to add the standard enthalpy value of $ CO $ and the standard enthalpy value of $ {H_2}O $
 $ \Delta {H_{reac\tan ts}} = \Delta {H_{CO}} + \Delta {H_{{H_2}O}} $
Now we can substitute the standard enthalpy value of the products and the standard enthalpy value of the reactants, to find the standard enthalpy of the reaction.
 $ \Delta {H_{reaction}} = \Delta {H_{products}} - \Delta {H_{reac\tan ts}} $
On substituting the values, we get,
 $
  \Delta {H_{reaction}} = \left[ {\left( { - 393.5kJ{{\left( {mol} \right)}^{ - 1}} \times 1molC{O_2}} \right) + \left( {0kJ{{\left( {mol} \right)}^{ - 1}} \times 1mol{H_2}} \right)} \right] - \\
  \left[ {\left( { - 110.5kJ{{\left( {mol} \right)}^{ - 1}} \times 1molCO} \right) + \left( { - 241.826kJ{{\left( {mol} \right)}^{ - 1}} \times 1mol{H_2}O} \right)} \right] \\
  $
 $ \Rightarrow \Delta {H_{reaction}} = \left[ {\left( { - 393.5kJ} \right) + \left( {0kJ} \right)} \right] - \left[ {\left( { - 110.5kJ} \right) + \left( { - 241.826kJ} \right)} \right] $
 $ \Rightarrow \Delta {H_{reaction}} = - 41.174kJ $
Therefore, the standard enthalpy of the reaction $ CO + {H_2}O \to C{O_2} + {H_2} $ is equal to $ - 41.174kJ $ .

Note:
The enthalpy of a system is influenced by a number of factors. Enthalpy is a broad concept that is influenced by the amount of material we deal with. The enthalpy value of a system is influenced by the state of reactants and materials (solid, liquid, or gas). The enthalpy value is affected by the reaction's path.