Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(s) + O₂ (g) ∆H° = 394 kJ/mol Determine the enthalpy for the reaction 2 SrCO₃ (s) → 2 Sr (s) + 2 C(s) + 3 O₂ (g).

The for the reaction is 72 kJ.

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 given chemical reaction follows:

The intermediate balanced chemical reaction are:

(1)    

(2)           ( × 2)

(3)         ( × 2)

The expression for enthalpy of the reaction follows:

Putting values in above equation, we get:

Hence, the for the reaction is 72 kJ.

c. is the answer

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