Question

Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine △ Hrxn for C(diamond) → C(graphite) with equations from the following list: (1) C(dianond)+O2(g)⟶CO2(g)ΔH=−395.4kJ (2) 2CO2(g)⟶2CO(g)+O2(g)ΔH=566.0kJ, (3) C(graphite)+O2(g)→CO2(g)ΔH=−393.5kJ, (4) 2CO(g)⟶C(graphite)+CO2(g)ΔH=−172.5kJ.

Answer 1

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1.9 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:

$C\text{ (diamond)}\rightarrow C\text{ (graphite)}$      $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C\text{ (diamond)}+O_2(g)\rightarrow CO_2(g)$    $\Delta H_1=-395.4kJ$

(2) $2CO_2(g)\rightarrow 2CO(g)+O_2(g)$     $\Delta H_2=566.0kJ$     ( × 2)

(3) $C\text{ (graphite)}+O_2(g)\rightarrow CO_2(g)$    $\Delta H_3=-393.5kJ$

(4) $2CO(g)\rightarrow 2C\text{ (graphite)}+CO_2(g)$     $\Delta H_4=-172.5kJ$     ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times \Delta H_2]+[1\times (\Delta H_3)]+[2\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-395.4))+(2\times (566.0))+(1\times (-393.5))+(2\times (-172.5))]=-1.9kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1.9 kJ.