consider the intermediate chemical reactions. ca(s)+co2(g)+1/2 o2(g)→caco3(s) δh1 = -812.8kj…

Explanation:

Step1: Write the target - equation and given equations

Target equation: $\ce{CaO(s) + CO2(g)\rightarrow CaCO3(s)}$ Given equations: Equation 1: $\ce{Ca(s)+CO2(g)+\frac{1}{2}O2(g)\rightarrow CaCO3(s)}$ $\Delta H_1=- 812.8\ kJ$ Equation 2: $\ce{2Ca(s)+O2(g)\rightarrow 2CaO(s)}$ $\Delta H_2=-1269.8\ kJ$

Step2: Manipulate the given equations to get the target equation

We want $\ce{CaO}$ on the left - hand side of the target equation. Equation 2 has $\ce{CaO}$ on the right - hand side. We need to reverse and halve Equation 2. When we reverse a chemical equation, the sign of the enthalpy change is reversed. When we multiply or divide an equation by a factor, the enthalpy change is also multiplied or divided by the same factor. Reversing and halving Equation 2 gives: $\ce{CaO(s)\rightarrow Ca(s)+\frac{1}{2}O2(s)}$, and the new $\Delta H$ for this equation is $\Delta H =+\frac{1269.8}{2}\ kJ$ Adding this new equation to Equation 1: $\ce{Ca(s)+CO2(g)+\frac{1}{2}O2(g)\rightarrow CaCO3(s)}$ $\Delta H_1=-812.8\ kJ$ $\ce{CaO(s)\rightarrow Ca(s)+\frac{1}{2}O2(s)}$ $\Delta H = +\frac{1269.8}{2}\ kJ$

$\ce{CaO(s)+CO2(g)\rightarrow CaCO3(s)}$

Answer:

The enthalpy of the second intermediate equation is halved and has its sign changed.