Question

Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ Express your answer numerically, to four significant figures and in terms of kJ.

Answer 1

Answer:

ΔH = 130.5 kJ

Explanation:

Hello,

In this case, by using the Hess law, we compute the enthalpy of the required reaction:

C(s) + H2O(g) --> CO(g) + H2(g)

Thus, the first step is to keep the following reaction unchanged:

C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ

Next, we invert and halve this reaction:

2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ

So the enthalpy of reaction is inverted and halved:

CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ

Then, we also invert and halve this reaction:

2 H2 (g) + O2 (g) → 2 H2O ΔH= -483.6 kJ

So the enthalpy of reaction is inverted and halved as well:

H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ

Finally, we add the three reactions to obtain the required reaction:

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + 1/2 O2 (g) + CO (g) + 1/2 O2 (g) + CO2 (g)

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)

= C (s) + H2O → H2 (g) CO (g)

So enthalpy is computed by:

ΔH = -393.5 kJ + 283 kJ + 241.8 kJ

ΔH = 130.5 kJ

Best regards.

Answer 2

Considering the Hess's Law, the enthalpy change for the reaction is 131.3 kJ.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C(s) + H₂O(g) → CO(g) + H₂(g)

You know the following reactions, with their corresponding enthalpies:

Equation 1: C (s) + O₂(g) → CO₂ (g)     ΔH = -393.5 kJ

Equation 2:  2 CO (g) + O₂ (g) → 2 CO₂ (g)     ΔH= -566.0 kJ

Equation 3: 2 H₂ (g) + O₂ (g) → 2 H₂O (g)     ΔH= -483.6 kJ  

• First step

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C(s) on reactant side and it is present in first equation so let's write this as such.

• Second step

Now, 1 mole of CO(g) must be a product and is present in the second equation. Since this equation has 2 moles of CO(g) on the reactant side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of CO(g).

When an equation is inverted, the sign of ΔH also changes.

And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also is divided by 2.

• Third step

Finally, 1 mole of H₂O(g) must be a reactant and is present in the third equation. Since this equation has 2 moles of CO(g) on the product side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of H₂O(g).

So, the sign of ΔH also changes and the variation of enthalpy is divided by 2.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: C (s) + O₂(g) → CO₂ (g)     ΔH = -393.5 kJ

Equation 2: CO₂ (g) → CO (g) + $\frac{1}{2}$ O₂ (g)     ΔH= 283 kJ

Equation 3: H₂O (g) → H₂ (g) + $\frac{1}{2}$ O₂ (g)    ΔH= 241.8 kJ  

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C(s) + H₂O(g) → CO(g) + H₂(g)  ΔH= 131.3 kJ

Finally, the enthalpy change for the reaction is 131.3 kJ.

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