Question

Using the standard enthalpies of formation, what is the standard enthalpy of reaction?
CO(g)+H2O(g)⟶CO2(g)+H2(g)
The formation values are as follows:

ΔHf (CO(g)) = −110.525KJ/mol
ΔHf (CO2(g)) = −393.509KJ/mol
ΔHf (H2O(g)) = −241.818KJ/mol
ΔHf (H2(g)) = 0
​

Answer 1

Answer:

Standard enthalpy for the given reaction is -41.166 kJ

Explanation:

Standard enthalpy of a reaction = $\sum [n_{i}\times \Delta H_{f}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}(reactant)_{j}]$

where $n_{i}$ and $n_{j}$ are number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence Standard enthalpy for the given reaction = $[(1\times \Delta H_{f}(CO_{2})_{g})]+[(1\times \Delta H_{f}(H_{2})_{g})]-[(1\times \Delta H_{f}(CO)_{g})]-[(1\times \Delta H_{f}(H_{2}O)_{g})]$

So, Standard enthalpy for the given reaction = $[(1\times -393.509]+[(1\times 0]-[(1\times -110.525]-[(1\times -241.818]$kJ = -41.166 kJ

Answer 2

The standard enthalpy of reaction from the given standard enthalpies of formation is;

∆Hrxn = -41.166 KJ/mol

The formula for standard enthalpy of reaction is;

∆Hrxn = Σ(n × ∆Hproducts) - Σ(n × ∆Hreactants)

Where n is the number of moles of the product or reactant as the case may be.

Now, the products in our reaction are;

CO2(g) and H2(g). We are given;

ΔHf (CO2(g)) = −393.509 KJ/mol

ΔHf (H2(g)) = 0

Thus;

Σ(n × ∆Hproducts) = (1 × −393.509) + (1 × 0)

Σ(n × ∆Hproducts) = -393.509 KJ/mol

Similarly, the reactants are; CO(g) and H2O(g). We are given;

ΔHf (CO(g)) = −110.525KJ/mol

ΔHf (H2O(g)) = −241.818KJ/mol

Thus;

Σ(n × ∆Hreactants) = (1 × -110.525) + (1 × -241.818)

Σ(n × ∆Hreactants) = −352.343 KJ/mol

Thus;

∆Hrxn = -393.509 - (−352.343)

∆Hrxn = -393.509 + 352.343

∆Hrxn = -41.166 KJ/mol

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