Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g)

Use the following:

A. –291.9 kJ
B. +291.9 kJ
C. –279.9 k
D. +279.9 kJ

D??

To calculate the enthalpy change (∆H) for the given reaction, we need to determine the difference in enthalpy between the products and reactants.

The enthalpy change can be calculated using the equation:

∆H = ΣH(products) - ΣH(reactants)

Looking at the balanced equation:
H2O(s) → H2(g) + 1/2O2(g)

The enthalpy change for the reaction can be determined using the enthalpy values for each substance.

The enthalpy change for 1 mole of H2O(s) is -285.8 kJ/mol.
The enthalpy change for 1 mole of H2(g) is 0 kJ/mol.
The enthalpy change for 1/2 mole of O2(g) is 0 kJ/mol.

Now, let's calculate the enthalpy change (∆H):

∆H = (0 kJ/mol + 0 kJ/mol) - (-285.8 kJ/mol)
∆H = 0 kJ/mol + 285.8 kJ/mol
∆H = 285.8 kJ/mol

Therefore, the enthalpy change (∆H) for the given reaction is +285.8 kJ/mol.

Based on the given options, none of them match the calculated value. Therefore, none of the options provided is correct.

To calculate the enthalpy change (∆H) for the given reaction, you need to use the enthalpy of formation values for each compound involved. The enthalpy of formation is the heat change that occurs when one mole of a compound is formed from its constituent elements, under standard conditions (25°C and 1 atm pressure).

The given reaction equation shows the formation of H2(g) and 1/2O2(g) from H2O(s). In order to calculate ∆H, we need to subtract the enthalpy of formation of the reactants from the enthalpy of formation of the products.

The enthalpy of formation value for H2O(s) is -285.8 kJ/mol (given), while the enthalpy of formation for H2(g) is 0 kJ/mol and for 1/2O2(g) is 0 kJ/mol (since they are the elemental forms).

Now, let's calculate the enthalpy change (∆H):

∆H = Σ Enthalpy of products - Σ Enthalpy of reactants
= [Enthalpy of H2(g) + Enthalpy of 1/2O2(g)] - Enthalpy of H2O(s)
= [0 kJ/mol + 0 kJ/mol] - (-285.8 kJ/mol)
= 0 kJ/mol + 285.8 kJ/mol
= 285.8 kJ/mol

Therefore, the enthalpy change (∆H) for the reaction is +285.8 kJ/mol. Comparing this result with the given options, the correct answer would be D. +279.9 kJ.

If we look at this the reverse direction.

H2(g) + 1/2O2(g) --> H2O(l) that's the heat formation liquid H2O in kJ/mol and my text shows that as -285.8 kJ/mol
Then H2O(l) ==> H2O(s) is another 6.0 kJ/mol to form ice and freezing water to ice is exothermic so that should be -6.0 kJ/mol. So I would go with -285.8 - 6 = ? Now you want the reverse of that so change the sign.