Question

Given the reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 890.5 kJ XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = + 221.1 kJ what is Δ H for this reaction? X ( s ) + 1 2 O 2 ( g ) + CO 2 ( g ) ⟶ XCO 3 ( s )

Answer 1

Answer: -1111.6 kJ

Step-by-step 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.

The given chemical reaction is,

`X(s)+12O_2(g)\rightarrow XO(s)`
`\Delta H_1=-890.5kJ` (1)

`XCO_3(s)\rightarrow XO(s)+CO_2(g)`
`\Delta H_2=+221.1kJ` (2)

Now we have to determine the value of
`\Delta H` for the following reaction i.e,

`X(s)+12O_2(g)+CO_2(g)\rightarrow XCO_3(s)`
`\Delta H_3=?` (3)

Subtracting (1) and (2) we get (3)

So, the value
`\Delta H_3` for the reaction will be:

`\Delta H_3=\Delta H_1-\Delta H_2=-890.5kJ-(+221.1kJ)=-1111.6kJ`

Hence, the value of
`\Delta H_3` for the reaction is -1111.6 kJ