Question

Question : Given the following data :

Ca(s) + 2C (graphite) $\stackrel{}{\mathbf{\to }}$CaC₂(s) ∆H = -62.8 kJ

Ca(s) + 2O₂(g) $\underset{}{\mathbf{\to }}$CaO(s) ∆H = - 635.5 kJ

CaO(s) + H₂O(l) role="math" localid="1648701987929" $\stackrel{}{\mathbf{\to }}$Ca〖(OH)〗_2(aq) ∆H = - 653.1 kJ

C₂H₂(g) +5/2O₂(g) $\underset{}{\mathbf{\to }}$ 2CO₂(g) +H₂O(l) ∆H = -1300. kJ

C(graphite) + O₂(g) $\stackrel{}{\mathbf{\to }}$CO₂(g) ∆H =- 393.5 kJ

Calculate ∆H for the reaction

CaC₂(s) + 2 H₂O(l) $\underset{}{\mathbf{\to }}$Ca(OH)₂(aq) + C₂H₂(g)

Short answer

Answer :

∆H for the reaction= - 712.8 kJ

Step-by-step solution

Reverse equation 1  

$C\mathrm{a}{C}_2\underset{}{\to }\mathrm{Ca}+2\mathrm{C}∆\mathrm{H}=62.8\mathrm{kJ}$

Reverse equation 4

$2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{C}}_2{\mathrm{H}}_2+\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{O}}_2∆\mathrm{H}=1300\mathrm{kJ}$

Multiply equation 5 by 2

$2\mathrm{C}+{\mathrm{O}}_2\underset{}{\to }{\mathrm{CO}}_2∆\mathrm{H}=2\times \left(-393.5\right)=-787.0\mathrm{kJ}$

Write all equations and solve them by the Hess law

${\mathrm{CaC}}_2\underset{}{\to }\mathrm{Ca}+2\mathrm{C}∆\mathrm{H}=62.8\mathrm{kJ}$

$$\begin{gathered} \mathrm{Ca}+2{\mathrm{O}}_2\underset{}{\to }\mathrm{CaO}∆\mathrm{H}=-635.5\mathrm{kJ} \\ \mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }\mathrm{Ca}{\left(\mathrm{OH}\right)}_2∆\mathrm{H}=-653.1\mathrm{kJ} \end{gathered}$$

$2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{C}}_2{\mathrm{H}}_2+\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{O}}_2∆\mathrm{H}=1300\mathrm{kJ}$

$2\mathrm{C}+{\mathrm{O}}_2\underset{}{\to }{\mathrm{CO}}_2∆\mathrm{H}=2\times \left(-393.5\right)=-787.0\mathrm{kJ}$

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${\mathrm{CaC}}_2+2{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+{\mathrm{C}}_2{\mathrm{H}}_2∆\mathrm{H}=-712.8\mathrm{kJ}$