Question

Calculate ΔHrxnΔHrxn for the following reaction: 3C(s)+4H2(g)→C3H8(l)3C(s)+4H2(g)→C3H8(l) Use the following reactions and given ΔHΔH values: C3H8(l)+5O2(g)→3CO2(g)+4H2O(g),ΔHC(s)+O2(g)→CO2(g),ΔH2H2(g)+O2(g)→2H2O(g),ΔH===−2026.6kJ−393.5kJ−483.5kJ

Answer 1

Answer: The $\Delta H^o_{rxn}$ for the reaction is -120.9 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$3C(s)+4H_2(g)\rightarrow C_3H_8(l)$      $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C_3H_8(l)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$    $\Delta H_1=-2026.6kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$     $\Delta H_2=-393.5kJ$     ( × 3)

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$     $\Delta H_2=-483.5kJ$     ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[3\times \Delta H_2]+[2\times \Delta H_3]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -(-2026.6))+(3\times (-393.5))+(2\times (-483.5))]=-120.9kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -120.9 kJ.

Answer 2

Considering the Hess's Law, the enthalpy change for the reaction is -120.9 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:

3 C(s) + 4 H₂(g) → C₃H₈(l)

You know the following reactions, with their corresponding enthalpies:

Equation 1: C₃H₈(l) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)     ΔH = –2026.6 kJ

Equation 2: C(s) + O₂(g) → CO₂(g)     ΔH = –393.5 kJ  

Equation 3: 2 H₂(g) + O₂(g) → 2 H₂O(g)    ΔH = -483.5 kJ

• First step

To obtain the enthalpy of the desired chemical reaction you need 1 mole of C₃H₈(l) on product side and it is present in first equation. Since this equation has 1 mole of C₃H₈(l) on the reactant side, it is necessary to locate 1 mole of C₃H₈(l) on the reactant side (invert it) When an equation is inverted, the sign of ΔH also changes.  

• Second step

Now, 2 moles of C(s) must be a reactant and is present in the second equation. Since this equation has 1 mole of C(s) on the reactant side, it is necessary to multiply it by 3. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiple by 3, the variation of enthalpy also is multiply by 3.

• Third step

Finally, 4 moles of H₂(g) must be a reactant and is present in the third equation. Since this equation has 2 mole of H₂(g) on the reactant side, it is necessary to multiply it by 2 and the variation of enthalpy also is multiply by 2.

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

Equation 1: 3 CO₂(g) + 4 H₂O(g) → C₃H₈(l) + 5 O₂(g)     ΔH = 2,026.6 kJ

Equation 2: 3 C(s) + 3 O₂(g) → 3 CO₂(g)     ΔH = –1,180.5 kJ  

Equation 3: 4 H₂(g) +  2 O₂(g) → 4 H₂O(g)    ΔH = -967 kJ

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

3 C(s) + 4 H₂(g) → C₃H₈(l)      ΔH= -120.9 kJ

Finally, the enthalpy change for the reaction is -120.9 kJ.

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