With a platinum catalyst, ammonia will burn in oxygen to give nitric oxide, NO.
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) ÄH = -906 kJ
What is the enthalpy change for the following reaction?
NO(g) + 3/2 H2O(g) NH3(g) + 5/4 O2(g)
To find the enthalpy change for the given reaction, we need to use the law of conservation of energy. This law states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, we will use the enthalpies of formation to calculate the enthalpy change.
The enthalpy change of a reaction can be calculated using the following equation:
ΔH = ∑(ΔHf(products)) - ∑(ΔHf(reactants))
Where ΔHf represents the standard enthalpy of formation of a substance.
To calculate the enthalpy change for the given reaction, we need to find the enthalpies of formation for each substance involved. The enthalpy of formation is defined as the enthalpy change for the formation of one mole of a compound from its elements in their most stable form at standard conditions (25°C and 1 atm pressure).
Using the given reaction, we can rearrange it to obtain the reactants and products in their elemental forms:
NH3(g) + 5/4 O2(g) 4 NO(g) + 6 H2O(g)
Now, let's calculate the enthalpy change using the enthalpies of formation. The enthalpy change can be expressed as:
ΔH = [4ΔHf(NO)] + [6ΔHf(H2O)] - [ΔHf(NH3)] - [(5/4)ΔHf(O2)]
To find the enthalpies of formation of the substances, we can use reference tables or thermodynamic databases. For this example, we'll assume the following values:
ΔHf(NO) = 90 kJ/mol
ΔHf(H2O) = -286 kJ/mol
ΔHf(NH3) = -46 kJ/mol
ΔHf(O2) = 0 kJ/mol
Plugging in the values, we get:
ΔH = [4(90 kJ/mol)] + [6(-286 kJ/mol)] - [(-46 kJ/mol)] - [(5/4)(0 kJ/mol)]
Simplifying the equation, we have:
ΔH = 360 kJ/mol - 1716 kJ/mol + 46 kJ/mol
Calculating the values, we get:
ΔH = -1310 kJ/mol
Therefore, the enthalpy change for the given reaction is -1310 kJ/mol.
To find the enthalpy change for the given reaction, we can use the fact that the enthalpy change for the forward reaction is the negative of the enthalpy change for the reverse reaction.
Given the enthalpy change for the forward reaction is -906 kJ, the enthalpy change for the reverse reaction will be +906 kJ.
Therefore, the enthalpy change for the reaction:
NO(g) + 3/2 H2O(g) NH3(g) + 5/4 O2(g)
is +906 kJ.
I assume that ÄH is supposed to be "delta H"
Note that the second reaction is the reverse of the first, but with the number of moles reacting 4 times less.
That should give you a pretty good hint of what the answer is.