Question

consider the intermediate chemical reactions.
ca(s)+co2(g)+1/2o2(g)→caco3(s) δh1 = - 812.8 kj
2ca(s)+o2(g)→2cao(s) δh2 = - 1,269.8 kj
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 is halved and has its sign changed.

Explanation:

Step1: Recall Hess's law

Hess's law states that the enthalpy change of an overall reaction is the sum of the enthalpy changes of the individual - reactions that make it up.

Step2: Manipulate the given reactions

The first reaction is \(Ca(s)+CO_{2}(g)+\frac{1}{2}O_{2}(g)
ightarrow CaCO_{3}(s)\) with \(\Delta H_1=- 812.8\ kJ\).
The second reaction is \(2Ca(s)+O_{2}(g)
ightarrow 2CaO(s)\) with \(\Delta H_2=-1269.8\ kJ\).
We want the overall reaction \(CaCO_{3}(s)
ightarrow CaO(s)+CO_{2}(g)\).
We need to reverse the first reaction (so \(\Delta H\) of the first - reversed reaction, \(\Delta H_{1r}=812.8\ kJ\)) and take half of the second reaction (so \(\Delta H\) of the halved second reaction, \(\Delta H_{2h}=\frac{-1269.8}{2}=-634.9\ kJ\)).

Step3: Calculate the overall enthalpy change

\(\Delta H_{overall}=\Delta H_{1r}+\Delta H_{2h}\).
\(\Delta H_{overall}=812.8+( - 634.9)=177.9\ kJ\).
When we reverse a reaction, the sign of its \(\Delta H\) is changed, and when we multiply or divide a reaction by a factor, the \(\Delta H\) is also multiplied or divided by the same factor.

Answer:

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