ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen monoxide, given that ΔH°f of NO2(g) is 33.90 kJ/mol.
dHrxn is -114.14 kJ/mol for what reaction?
To calculate ΔH°f of gaseous nitrogen monoxide (NO(g)), we can use the given information about the enthalpy change of reaction (ΔH°rxn) and the enthalpy of formation of nitrogen dioxide (ΔH°f of NO2(g)).
The enthalpy change of reaction (ΔH°rxn) can be related to the enthalpy of formation (ΔH°f) by the following equation:
ΔH°rxn = ΣnΔH°f(products) - ΣmΔH°f(reactants)
In this case, the equation becomes:
-114.14 kJ/mol = ΔH°f(NO(g)) - ΔH°f(NO2(g))
We are given the value of ΔH°f(NO2(g)) as 33.90 kJ/mol, so we can plug this value into the equation:
-114.14 kJ/mol = ΔH°f(NO(g)) - 33.90 kJ/mol
To solve for ΔH°f(NO(g)), we rearrange the equation:
ΔH°f(NO(g)) = -114.14 kJ/mol + 33.90 kJ/mol
Calculating the sum, we have:
ΔH°f(NO(g)) = -80.24 kJ/mol
Therefore, the ΔH°f of gaseous nitrogen monoxide (NO(g)) is approximately -80.24 kJ/mol.
To calculate ΔH°f of gaseous nitrogen monoxide (NO(g)), we need to use the thermochemical equation and the fact that ΔH°rxn is -114.14 kJ/mol.
The given thermochemical equation is:
2NO(g) + O2(g) → 2NO2(g)
We know that the ΔH°rxn for this reaction is -114.14 kJ/mol. However, we need to find the ΔH°f for NO(g) first.
ΔH°rxn is the enthalpy change for the reaction as written, and it is equal to the sum of the enthalpy changes for the formation of the products minus the enthalpy changes for the formation of the reactants.
ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)
Let's start by finding the enthalpy change for the formation of NO2(g) using the given value:
ΔH°f(NO2(g)) = 33.90 kJ/mol
The stoichiometric coefficient for NO2(g) is 2 in the balanced equation. So, we need to divide the given value by 2 to find the enthalpy change for the formation of 1 mole of NO2(g) from its elements:
ΔH°f(NO2(g)) = 33.90 kJ/mol / 2 = 16.95 kJ/mol
Now, we can substitute the enthalpy changes for NO2(g) and NO(g) into the thermochemical equation and solve for ΔH°f(NO(g)):
-114.14 kJ/mol = 2(16.95 kJ/mol) + ΔH°f(NO(g))
-114.14 kJ/mol = 33.9 kJ/mol + ΔH°f(NO(g))
Subtracting 33.9 kJ/mol from both sides:
-148.04 kJ/mol = ΔH°f(NO(g))
Therefore, the ΔH°f of gaseous nitrogen monoxide (NO(g)) is -148.04 kJ/mol.
Note: The negative sign indicates that the formation of gaseous nitrogen monoxide from its elements is an exothermic process, meaning that it releases energy.