If we have :
H2(g) + 1/2 O2(g) -> H2O(l) dHf= -285.5 Kj
N2O5(g) + H2O(l) -> 2HNO3 dHf = -76.6 Kj
1/2 N2 + 3/2 O2 + 1/2 H2 -> HNO3 dHf = - 174.1 Kj
then , what is the dHf of
2N2 + 5O2 -> 2N2O5
Reverse equation 3 and add to the reverse of equation 2 and add twice equation 3. When you get the final equation it will be 1/2 of what you want; just multiply that by 2.
When reversing equations change the sign of dH. When multiplying equations, multiply the dH value by the multiplier.
I'm confused , can you do the solution please ?
To find the enthalpy change of the reaction 2N2 + 5O2 -> 2N2O5, we can use the given enthalpies of formation (dHf) for the reactants and products involved. The enthalpy change (dH) of a reaction can be calculated using the following equation:
dH = Σn(dHf products) - Σn(dHf reactants)
Where:
dH = enthalpy change
Σn = sum of coefficients
dHf = standard enthalpy of formation
Let's break down the reaction into the sum of reactants and products:
Reactants:
2N2
5O2
Products:
2N2O5
Now, we can use the enthalpies of formation given for each species to calculate the enthalpy change:
For the reactants:
2N2: dHf = 0 (since it is the standard state)
5O2: dHf = 0 (since it is the standard state)
For the products:
2N2O5: dHf = ?
Using the given dHf values for H2O and HNO3, we can calculate the enthalpy change for the products:
2HNO3: dHf = 2(-76.6 Kj) = -153.2 Kj
Now, we can substitute all the values into the enthalpy change equation:
dH = Σn(dHf products) - Σn(dHf reactants)
= (2)(-153.2 Kj) - (2)(0 Kj) - (5)(0 Kj)
= -306.4 Kj
Therefore, the enthalpy change (dH) for the reaction 2N2 + 5O2 -> 2N2O5 is -306.4 Kj.