Br2 == 2Br
24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2
=> partial pressure of Br, P Br = 0.24 bar, and
partical pressure of Br2, P Br2 = 0.76 bar
kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
Looking at this equation P= (pa*pb)/ (pa+(pb-pa)) ya where pa=vap press a and ya= vap composition a and P= total pressure,it relates vapor pressure mixture to vapor composition. This is derived using the combination of Dalton's and Raoult's laws.
Answer 1) : The density of the hot air inside the balloon can be found out by using ideal gas equation;
PV = nRT;
As n is number of moles and in gases, number of moles along with mass per mole is equal to the density of the gas.
If the moles in the gas are more the density will be more.
here, density (ρ) = mass (m) / volume (V); substituting in the ideal gas equation we get,
ρ = mP / RT
Answer 2) ρ (hot air) = ρ (cold air) X 
Here according to the formula because T(hot air) >T(cold air),
So, the density of hot air greater than the density of cold air.
The relationship between the ρ (h) = ρ(c) X
<span>According to the law of conservation of energy and due that all the chemical energy is converted to other three types of energy, the total sum of these three energies after the explosion must be the same than the initial energy, that is 100 units.</span>
