Question

Given the reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 505.9 kJ XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = + 199.3 kJ what is Δ H for this reaction? X ( s ) + 1 2 O 2 ( g ) + CO 2 ( g ) ⟶ XCO 3 ( s )

Answer 1

Answer: -705.2 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

$X(s)+12O_2(g)\rightarrow XO(s)$    $\Delta H^0_1=-505.9kJ$   (1)

$XCO_3(s)\rightarrow XO(s)+CO_2(g)$ $\Delta H^0_2=+199.3kJ$  (2)

The final reaction is:  

$X(s)+12O_2(g)+CO_2(g)\rightarrow XCO_3(s)$  $\Delta H^0_3=?$   (3)

Reversing (2)

$XO(s)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^0_2'=-199.3kJ$  (2')

By adding (1) and (2')

$\Delta H^0_3=\Delta H^0_1+\Delta H^0_2'=-505.9kJ+(-199.3kJ)=-705.2kJ$

Hence $\Delta H^0=-705.2kJ$.