if enthalpies for formation of p4o10, h2o and h3po4 are -2984, -285.9 and -1279 respectively calculate the enthalpy change using hess's law
To calculate the enthalpy change using Hess's law, you need to consider the balanced chemical equation and determine the overall reaction using the given enthalpies of formation for individual reactions.
First, let's start by writing the balanced equation for the reaction:
P4O10 + 6H2O ⟶ 4H3PO4
Now, let's analyze the given reactions and their enthalpies of formation, and see how they can be combined to give the overall reaction:
1) 4P + 5O2 ⟶ P4O10 (Enthalpy of formation = -2984 kJ/mol)
2) H2 + 1/2O2 ⟶ H2O (Enthalpy of formation = -285.9 kJ/mol)
3) P4 + 6H2 ⟶ 4H3PO4 (Enthalpy of formation = -1279 kJ/mol)
Since we need 4 moles of H3PO4, we can multiply reaction (3) by 4 to balance the number of moles of H3PO4:
4[P4 + 6H2 ⟶ 4H3PO4] gives us:
4P4 + 24H2 ⟶ 16H3PO4
Next, we rearrange the equations to match the desired overall equation (P4O10 + 6H2O ⟶ 4H3PO4):
Multiplying equation (1) by 4 to get the same number of moles of P4O10:
4[P4 + 5O2 ⟶ P4O10] gives us:
4P4 + 20O2 ⟶ 4P4O10
Now, let's manipulate equation (2) to obtain the same number of moles of H2O as the desired overall equation. For this, we need to multiply it by 12:
12[H2 + 1/2O2 ⟶ H2O] gives us:
12H2 + 6O2 ⟶ 12H2O
Now, sum up all the manipulated equations to get the desired reaction:
4P4 + 20O2 + 12H2 + 6O2 ⟶ 4P4O10 + 12H2O + 16H3PO4
Finally, let's sum up the enthalpies of formation for the manipulated equations to obtain the enthalpy change:
Enthalpy change = (4 x -2984) + (12 x -285.9) + (16 x -1279) - (4 x 0) - (12 x 0) - (20 x 0) - (6 x 0) = -55181.6 kJ
Therefore, the enthalpy change using Hess's law for the given reaction is -55181.6 kJ.
Two or three points here.
1. You can't get lazy in chemistry and not use the caps key. Those symbols don't mean anything to me.
2. -2984, -285.9, -1279 what units.
3. Calculate enthalpy change for what?