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tabbydory3366
24.02.2020 •
Chemistry
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ?Given:PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = +157 kJP4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ
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Ответ:
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![\bigtriangledown H^\circ_{rxn}](/tpl/images/0521/1764/88805.png)
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
Ответ:
The air pressure at sea level is about 14.7 pounds per square inch (psi). For every foot above sea level, the air pressure drops about 0.0005263 psi. According to this information, at about what height above sea level is the air pressure 13.2 psi?