Given H2(g) + (1/2)O2(g) ---> H2O(l), dH = -286 kJ/mol, determine the standard enthalpy change for the reaction 2h2O(l) ---> 2H2(g) + O2(g)
2H2O(l) ---> 2H2(g) + O2(g)
H2(g) + (1/2) O29g) ---> H2O (l) : dH = -286 kJ/mols
2H2O(l) ---> 2H2(g) + 2O2(g) : would dH be -572 or +572?
You want twice the reverse reaction given in the problem. So that's -(-286)*2 = ?
To determine the standard enthalpy change for the reaction 2H2O(l) ---> 2H2(g) + O2(g), we can use the information given about the reaction H2(g) + (1/2)O2(g) ---> H2O(l), where the standard enthalpy change (dH) is -286 kJ/mol.
Since the given reaction contains 2 moles of water (H2O), we can multiply the enthalpy change of the first reaction by 2:
2H2(g) + 2(1/2)O2(g) ---> 2H2O(l)
Therefore, the standard enthalpy change for this reaction would be -286 kJ/mol * 2 = -572 kJ/mol.
So, the answer is -572 kJ/mol.
To determine the standard enthalpy change for the reaction:
2H2O(l) ---> 2H2(g) + O2(g)
First, let's look at the given reaction:
H2(g) + (1/2)O2(g) ---> H2O(l) : dH = -286 kJ/mol
Notice that this equation is the reverse of the given reaction. The enthalpy change of the reverse reaction is the negative of the original reaction, so the enthalpy change for the reverse reaction is +286 kJ/mol.
Now, let's rewrite the given reaction as two separate reactions:
H2O(l) ---> H2(g) + (1/2)O2(g)
H2O(l) ---> H2(g) + (1/2)O2(g) : dH = +286 kJ/mol
Since we want to find the enthalpy change for 2H2O(l), we can double the enthalpy change for the first reaction:
2H2O(l) ---> 2H2(g) + O2(g) : dH = +572 kJ/mol
So, the standard enthalpy change for the reaction 2H2O(l) ---> 2H2(g) + O2(g) is +572 kJ/mol.