Question

Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one half upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s). Delta H 1 equals negative 812.8 kilojoules. Second: 2 upper C a (s) plus upper O subscript 2 (g) right arrow 2 upper C a upper O (s). Delta H 2 equals negative 1, 269 kilojoules. The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (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. is halved. has its sign changed. is unchanged.

Answer 1

Answer: 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:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$     $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$    $\Delta H_1=-812.8kJ$  

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$     $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.