Question

Given the reactions: X(s) + 1/2 O2(g) ⟶ XO(s) ΔH = -817.3 kJ XCO3(s) ⟶ XO(s) + CO2(g) ΔH = +491.7 kJ What is the ΔH for the reaction: X(s) + 1/2 O2(g) + CO2(g) ⟶ XCO3(s)

Answer 1

By using Hess's Law to combine the given reactions, reversing the second reaction, and summing their enthalpy changes, the enthalpy change (ΔH) for the reaction X(s) + 1/2 O2(g) + CO2(g) → XCO3(s) is found to be -1309.0 kJ.

Step-by-step explanation:

The student is asking for the enthalpy change (ΔH) for the reaction using Hess's Law and the provided reactions. To find this, we need to manipulate the given reactions to arrive at the target equation. The first reaction is:

X(s) + 1/2 O2(g) → XO(s) with ΔH = -817.3 kJ

The second reaction is:

XCO3(s) → XO(s) + CO2(g) with ΔH = +491.7 kJ

To solve this, we reverse the second equation and then add it to the first, taking care to reverse the sign of ΔH for the second reaction:

XO(s) + CO2(g) → XCO3(s) with ΔH = -491.7 kJ

Adding the two reactions, we get:

X(s) + 1/2 O2(g) + CO2(g) → XCO3(s)

By adding the ΔH values of the two reactions, -817.3 kJ and -491.7 kJ, we find that the enthalpy change for the target reaction is ΔH = -1309.0 kJ.