If the standard enthalpy of formation of PH3(g) is +5.4 kJ mol-1, calculate the reaction enthalpy for
2PH3(g)--->2P(s)+3H2(g)
Sam--All of these are about alike. Drum that dHrxn = (n*dHf products) - (n*dHf reactants) into your head.
To calculate the reaction enthalpy, we need to use the standard enthalpies of formation of the reactants and products. The standard enthalpy of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its elements, where all substances are in their standard states at a given temperature and pressure.
Given:
ΔHf (PH3(g)) = +5.4 kJ mol-1
We can write the balanced equation for the reaction:
2PH3(g) ---> 2P(s) + 3H2(g)
Now, let's calculate the reaction enthalpy (ΔH) using the standard enthalpies of formation:
ΔH = (sum of the standard enthalpies of formation of products) - (sum of the standard enthalpies of formation of reactants)
The given balanced equation shows that we have 2 moles of PH3 forming, 2 moles of P, and 3 moles of H2.
Reactants:
2PH3(g): Since the reaction enthalpy refers to the formation reaction, we will use the standard enthalpy of formation of PH3(g) for each mole of PH3.
ΔHf (PH3(g)) = +5.4 kJ mol-1
Products:
2P(s): The standard enthalpy of formation of an element in its standard state is defined as zero.
ΔHf (P(s)) = 0 kJ mol-1
3H2(g): We will use the standard enthalpy of formation of H2(g) for each mole of H2.
ΔHf (H2(g)) = 0 kJ mol-1 (As H2(g) is the standard state for hydrogen)
Now let's calculate the reaction enthalpy (ΔH):
ΔH = (2P(s) + 3H2(g)) - (2PH3(g))
ΔH = 0 + 0 - (2 * 5.4)
ΔH = -10.8 kJ mol-1
Therefore, the reaction enthalpy for the given reaction 2PH3(g) ---> 2P(s) + 3H2(g) is -10.8 kJ mol-1.