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Question: What is the standard enthalpy of the reaction \( CO + {H_2}O \to C{O_2} + {H_2} \) ?...

What is the standard enthalpy of the reaction CO+H2O→CO2+H2CO + {H_2}O \to C{O_2} + {H_2} ?

Explanation

Solution

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 COCO is −110.5kJ(mol)−1- 110.5kJ{\left( {mol} \right)^{ - 1}}
The standard enthalpy value of H2O{H_2}O is −241.826kJ(mol)−1- 241.826kJ{\left( {mol} \right)^{ - 1}}
The standard enthalpy value of CO2C{O_2} is −393.5kJ(mol)−1- 393.5kJ{\left( {mol} \right)^{ - 1}}
The standard enthalpy value of H2{H_2} is 0kJ(mol)−10kJ{\left( {mol} \right)^{ - 1}}
Once we have obtained these numbers, put them into the standard enthalpy formation equation and solve for the standard enthalpy.
ΔHreaction=ΔHproducts−ΔHreactan⁡ts\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 CO2C{O_2} and the standard enthalpy value of H2{H_2}
ΔHproducts=ΔHCO2+ΔHH2\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 COCO and the standard enthalpy value of H2O{H_2}O
ΔHreactan⁡ts=ΔHCO+ΔHH2O\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.
ΔHreaction=ΔHproducts−ΔHreactan⁡ts\Delta {H_{reaction}} = \Delta {H_{products}} - \Delta {H_{reac\tan ts}}
On substituting the values, we get,
ΔHreaction=[(−393.5kJ(mol)−1×1molCO2)+(0kJ(mol)−1×1molH2)]− [(−110.5kJ(mol)−1×1molCO)+(−241.826kJ(mol)−1×1molH2O)]  \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] \\\
⇒ΔHreaction=[(−393.5kJ)+(0kJ)]−[(−110.5kJ)+(−241.826kJ)]\Rightarrow \Delta {H_{reaction}} = \left[ {\left( { - 393.5kJ} \right) + \left( {0kJ} \right)} \right] - \left[ {\left( { - 110.5kJ} \right) + \left( { - 241.826kJ} \right)} \right]
⇒ΔHreaction=−41.174kJ\Rightarrow \Delta {H_{reaction}} = - 41.174kJ
Therefore, the standard enthalpy of the reaction CO+H2O→CO2+H2CO + {H_2}O \to C{O_2} + {H_2} is equal to −41.174kJ- 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.