i have one more question:
I know this one has to do with hess's law but i'm not sure of how to get the enthalpy of vaporization from this?
Clculate the enthalpy of vaporization of solid potassium bromide to a gas of ions, the process KBr--->K{+1}(g)+Br{-1}(g), from the following information and the enthalpy of formation of KBr(s), which is -394kJ/mol
Atomization of K: K(s)--->K(g) delta H given
Ionization of K: K(g)--> K{+1}(g)+ e(g) delta H given
VAporization of bromine: Br2(l)-->Br2(g) delta H given
Dissociation of bromine: Br2(g)-->2Br(g) delta H given
Electron attachment of bromine: Br(g)+e--->Br{-1}(g) delta H given
I just posted it at the original spot below.
thanx so much!
To calculate the enthalpy of vaporization of solid potassium bromide to a gas of ions (K{+1}(g) + Br{-1}(g)), you can use Hess's law.
Hess's law states that the total enthalpy change for a chemical reaction is independent of the pathway between the initial and final states. In other words, if you can express the reaction you want in terms of a series of known reactions, you can sum up the enthalpy changes of those reactions to obtain the enthalpy change of the desired reaction.
Here's how you can apply Hess's law to calculate the enthalpy of vaporization of potassium bromide:
1. Write down the overall reaction you want to calculate: KBr(s) → K{+1}(g) + Br{-1}(g)
2. Identify a series of known reactions that can be combined to give the desired reaction. In this case, we can break down the process into the following steps:
a) Atomization of potassium: K(s) → K(g)
b) Ionization of potassium: K(g) → K{+1}(g) + e(g)
c) Vaporization of bromine: Br2(l) → Br2(g)
d) Dissociation of bromine: Br2(g) → 2Br(g)
e) Electron attachment of bromine: Br(g) + e → Br{-1}(g)
3. Determine the enthalpy changes for each known reaction. These enthalpy changes are given in the problem statement.
∆H for atomization of K = ∆H1
∆H for ionization of K = ∆H2
∆H for vaporization of Br2 = ∆H3
∆H for dissociation of Br2 = ∆H4
∆H for electron attachment of Br = ∆H5
4. Apply Hess's law by combining the known reactions to obtain the desired reaction:
KBr(s) → K{+1}(g) + Br{-1}(g)
Reaction 1: K(s) → K(g) (reverse sign, -∆H1)
Reaction 2: K(g) + Br2(g) → K{+1}(g) + Br2(g) + e(g) (reverse sign, -∆H2)
Reaction 3: Br2(l) → Br2(g) (+∆H3)
Reaction 4: Br2(g) → 2Br(g) (+∆H4)
Reaction 5: Br(g) + e → Br{-1}(g) (+∆H5)
Adding up these reactions cancels out the intermediate species (gases) and gives the desired reaction:
KBr(s) → K{+1}(g) + Br{-1}(g)
The enthalpy change for this reaction is:
∆H(total) = - ∆H1 - ∆H2 + ∆H3 + ∆H4 + ∆H5
5. Substitute the given values for the enthalpy changes and calculate the total enthalpy change (∆H(total)).
∆H(total) = - ∆H1 - ∆H2 + ∆H3 + ∆H4 + ∆H5
Finally, substitute the known values and calculate ∆H(total) to obtain the enthalpy of vaporization of solid potassium bromide to a gas of ions.