Calculate the enthalpy of the reaction of boron trioxide with steam: B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g)
Given:
H2O(l)
H2(g) + 1⁄2 O2(g)
2B(s) + 3H2(g)
2B(s) + 3/2 O2(g) → B2O3(s)
→ H2O(g) → H2O(l) → B2H6(g)
44 kJ/mol -286 kJ/mol 36 kJ/mol -1273 kJ/mol
dHrxn = (n*dHf products) - (n*dHf reactants)
To calculate the enthalpy of the reaction, we need to use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps.
Given:
H2O(l) → H2(g) + 1⁄2 O2(g) : ΔH = 44 kJ/mol
2B(s) + 3H2(g) → 2B(s) + 3/2 O2(g) : ΔH = -1273 kJ/mol
We want to find the enthalpy of the reaction:
B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g)
Step 1: Convert B2O3(s) to B(s)
B2O3(s) → 2B(s) + 3/2 O2(g) : ΔH = -1273 kJ/mol (Given)
Step 2: Convert H2O(g) to H2O(l)
H2O(g) → H2O(l) : ΔH = 0 kJ/mol (Since there is no change in state)
Step 3: Convert H2O(l) to H2(g) + 1/2 O2(g)
H2O(l) → H2(g) + 1/2 O2(g) : ΔH = -44 kJ/mol (In the reverse direction, so the sign is opposite -44 kJ/mol)
Step 4: Convert 2B(s) + 3/2 O2(g) to B2O3(s)
2B(s) + 3/2 O2(g) → B2O3(s) : ΔH = +1273 kJ/mol (Since this is the reverse of the first step, the sign is opposite +1273 kJ/mol)
Step 5: Convert B2O3(s) to 3O2(g) + B2H6(g)
B2O3(s) → 3O2(g) + B2H6(g) : ΔH = ΔH of the reaction we want to find
To calculate the enthalpy of the reaction we want, we add the enthalpy changes of the individual steps:
ΔH = (-1273 kJ/mol) + (0 kJ/mol) + (-44 kJ/mol) + (1273 kJ/mol)
ΔH = -44 kJ/mol
Therefore, the enthalpy change of the reaction of boron trioxide with steam is -44 kJ/mol.
To calculate the enthalpy of the reaction, you will use the concept of Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of individual reactions.
1. Identify the reactions in the given information and their respective enthalpy changes:
a) 2B(s) + 3/2 O2(g) → B2O3(s) - Enthalpy change: -1273 kJ/mol
b) 3H2(g) + 1.5O2(g) → 3H2O(g) - Enthalpy change: 286 kJ/mol
c) B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g) - Enthalpy change: ?
2. Rearrange the equations to match the desired reaction:
a) 2B(s) + 3/2 O2(g) → B2O3(s) - Multiply this equation by 2.
Resulting equation: 4B(s) + 3O2(g) → 2B2O3(s)
b) 3H2(g) + 1.5O2(g) → 3H2O(g) - Multiply this equation by 2/3.
Resulting equation: 2H2(g) + O2(g) → 2/3 H2O(g)
3. Flip the sign of equation b) as it needs to be reversed in the final reaction:
Revised equation: 2/3 H2O(g) → 2H2(g) + O2(g) - Enthalpy change: -286 kJ/mol
4. Add the equations together while canceling out the common compounds:
4B(s) + 3O2(g) + 2/3 H2O(g) → 2B2O3(s) + 2H2(g) + O2(g) - Enthalpy change: ?
5. Simplify the equation by removing the common compounds:
4B(s) + 2/3 H2O(g) → 2B2O3(s) + 2H2(g) - Enthalpy change: ?
6. The final equation matches the desired reaction, and since we are interested in the enthalpy change of the reaction:
Enthalpy change of the desired reaction: -1273 kJ/mol + (-286 kJ/mol) = -1559 kJ/mol
Therefore, the enthalpy change of the reaction of boron trioxide with steam is -1559 kJ/mol.