Ca(OH)2(s) −→ CaO(s) + H2O(g)
Given:
Ca(OH)2(s) −→ Ca(s) + O2(g) + H2(g)
∆H = 973.3 kJ/mol
2 Ca(s) + O2(g) −→ 2 CaO(s)
∆H = −1269.8 kJ/mol
2 H2(g) + O2(g) −→ 2 H2O(g)
∆H = −483.6 kJ/mol
Answer in units of kJ/mol.
To find the enthalpy change for the reaction:
Ca(OH)2(s) → CaO(s) + H2O(g)
We need to use the given enthalpy changes of the three reactions provided. However, the given enthalpy changes must be manipulated and combined to obtain the desired reaction.
Given reactions:
1) Ca(OH)2(s) → Ca(s) + O2(g) + H2(g) ∆H = 973.3 kJ/mol
2) 2 Ca(s) + O2(g) → 2 CaO(s) ∆H = -1269.8 kJ/mol
3) 2 H2(g) + O2(g) → 2 H2O(g) ∆H = -483.6 kJ/mol
Step 1:
Since we want to find the enthalpy change for Ca(OH)2(s) → CaO(s) + H2O(g), we need to make the coefficients in the given reactions align with the desired reaction.
Looking at the given reactions, we notice that reaction 1 has Ca(OH)2 and H2O in it, which matches the desired reactants. We can use this reaction and modify it to achieve our desired reaction.
Multiply reaction 1 by 2:
2 Ca(OH)2(s) → 2 Ca(s) + 2 O2(g) + 2 H2(g) ∆H = 2 * 973.3 = 1946.6 kJ/mol
Step 2:
The goal now is to manipulate the first equation and combine it with equations 2 and 3 such that the overall equation matches our desired reaction.
Add equation 2 and equation 3 to the modified equation 1:
2 Ca(OH)2(s) + 2 Ca(s) + 2 O2(g) + 2 H2(g) → 2 Ca(s) + 2 O2(g) + 2 H2O(g) ∆H = 1946.6 - 1269.8 - 483.6 = 193 kJ/mol
Step 3:
Cancel out the common species on both sides of the equation:
2 Ca(OH)2(s) + 2 H2(g) → 2 H2O(g) ∆H = 193 kJ/mol
Finally, divide the coefficients throughout the equation by 2 to simplify it:
Ca(OH)2(s) + H2(g) → H2O(g) ∆H = 193 / 2 = 96.5 kJ/mol
Therefore, the enthalpy change for the reaction Ca(OH)2(s) → CaO(s) + H2O(g) is 96.5 kJ/mol.