Given the reactions, x(s)+12o2(g)⟶xo(s) δ=−549.3 kj x(s)+12o2(g)⟶xo(s) δh=−549.3 kj xco3(s)⟶xo(s)+co2(g)δ=+170.9 kj xco3(s)⟶xo(s)+co2(g)δh=+170.9 kj what is δδh for this reaction? x(s)+12o2(g)+co2(g)⟶xco3(s)

-720.2 kJ

Explanation:

We have two equations:

(I)   X + ½O₂ ⟶ XO;       ΔH = −549.3 kJ

(II) XCO₃ ⟶ XO + CO₂; ΔH =  +170.9 kJ  

From these, we must devise the target equation:

(III) X +½O₂ +CO₂ ⟶ XCO₃; ΔH = ?

The target equation has X on the left, so you rewrite Equation(I).

(I) X + ½O₂ ⟶ XO; ΔH = −549.3 kJ

Equation (IV) has XO on the right, and that is not in the target equation.

You need an equation with XO on the left, so you reverse Equation (II).  

When you reverse an equation, you reverse the sign of its ΔH.  

(IV)  XO + CO₂ ⟶ XCO₃ ;  ΔH = -170.9 kJ  

Now, you add equations (I) and (IV), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation (III):

(I)   X + ½O₂ ⟶ XO;                ΔH = -549.3 kJ

(IV) XO + CO₂ ⟶ XCO₃ ;       ΔH =  -170.9 kJ  

(III) X + ½O₂ + CO₂ ⟶ XCO₃; ΔH = -720.2 kJ

the dependant variable is spilled water

the independent is the heat