A sample of gas is held at constant pressure. Increasing the Kelvin temperature of this gas sample causes the average kinetic en
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Answer:
The net energy is 2.196 eV
Explanation:
Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.
Using:
ΔE = h*c*(1/λ - 1/λ
)
where:
ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602* J.
h = 4.135* eVs
c = 3* m/s
λ = 300 nm = 300*
m
λ = 640 nm = 640*
m
Thus:
ΔE = 4.135* eVs*3*
m/s*(
)
ΔE = 4.135**3*
*1.77*
eV = 2.196 eV
Answer:
Therefore = -1835 KJ
Explanation:
Enthalpy is denoted by H.
Enthalpy: Total heat change in a chemical reaction is called enthalpy.
The change of entalpy of a reaction is denoted by
Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.
g= gas
S= solid
P₄(g)+10Cl₂(g)→ 4Cl₅(s) =?
PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) = +157KJ
P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) = -1207 KJ
If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.
We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.
PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) = -157KJ
Multiplying 4 with equation (3)
4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) =4×( -157)KJ= -628 KJ
Adding equation (2) and (4) we get
P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) =( -1207-628)KJ
⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) = - 1835KJ
⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) = -1835 KJ
Therefore = -1835 KJ