Given the following data $$\mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) \longrightarrow \mathrm{CaC}_{2}(s)$$ \(\Delta H=-62.8 \mathrm{kJ}\) $$ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CaO}(s) $$ \(\Delta H=-635.5 \mathrm{kJ}\) $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) $$ \(\Delta H=-653.1 \mathrm{kJ}\) $$ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1300 . \mathrm{kJ}\) $$ \mathrm{C}(\text {graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) $$ \(\Delta H=-393.5 \mathrm{kJ}\) calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

The target reaction is $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$ Our goal is to find its enthalpy change, ∆H.

We need to manipulate the given reactions in such a way that they will sum up to the target reaction. The most straightforward way we can do that is to find a way for the given reactions to cancel out as many compounds as possible while resulting in the target reaction. Here's how we can manipulate them: 1. Reverse the first reaction and double the second and fifth reaction to obtain the correct coefficients for the target reaction: $$\mathrm{CaC}_{2(s)} \longrightarrow \mathrm{Ca(s)} + 2 \mathrm{C(graphite)}$$ \(\Delta H = 62.8\,\mathrm{kJ}\) (reversed) $$2(\mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CaO}(s))$$ \(2 \Delta H = 2(-635.5)\,\mathrm{kJ}\) (multiplied by 2) $$2(\mathrm{C}(\text {graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm(g)))$$ \(2 \Delta H = 2(-393.5)\,\mathrm{kJ}\) (multiplied by 2) 2. Add the third and fourth reaction as they are: $$\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)$$ \(\Delta H = -653.1\,\mathrm{kJ}\) $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ \(\Delta H = -1300\,\mathrm{kJ}\)

Now, sum up the manipulated reactions. The intermediate compounds will cancel out, resulting in the target reaction: $$ \mathrm{CaC}_{2}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

Add the enthalpies change (∆H) of the manipulated reactions to find the ∆H of the target reaction: $$\Delta H_\text{target} = 62.8\,\mathrm{kJ} + 2(-635.5)\,\mathrm{kJ} + 2(-393.5)\,\mathrm{kJ} - 653.1\,\mathrm{kJ} - 1300\,\mathrm{kJ}$$ $$\Delta H_\text{target} = -1799.9\,\mathrm{kJ}$$ So the enthalpy change, ∆H, for the target reaction is -1799.9 kJ.