Question

consider these intermediate chemical equations. ca(s) + co₂(g) + 1/2o₂(g) → caco₃(s) δh₁ = -812.8kj 2ca(s) + o₂(g) → 2cao(s) δh₂ = -1,269.8kj the final overall chemical equation is cao(s) + co₂(g) → caco₃(s). when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation has its sign changed. is unchanged. is halved. is halved and has its sign changed.

Explanation:

Step1: Manipulate the intermediate equations to get the overall equation

We want to get $CaO(s)+CO_2(g)\to CaCO_3(s)$ from the given intermediate equations:
Equation 1: $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\to CaCO_3(s)\quad \Delta H_1 = - 812.8kJ$
Equation 2: $2Ca(s)+O_2(g)\to 2CaO(s)\quad \Delta H_2=-1269.8kJ$
We need to reverse and halve the second - equation to get $CaO(s)\to Ca(s)+\frac{1}{2}O_2(g)$. When we reverse a chemical equation, the sign of the enthalpy change is reversed, and when we multiply or divide an equation by a factor, the enthalpy change is multiplied or divided by the same factor.

Step2: Analyze the change in $\Delta H_2$

When we reverse and halve the second equation, the new $\Delta H$ for the modified second equation is $\Delta H=\frac{1}{2}\times(-\Delta H_2)$. So the enthalpy of the second intermediate equation has its sign changed and is halved.

Answer:

is halved and has its sign changed.