If the moles and volume of a gas are held constant, dropping the temperature from 40 degrees Celsius to 20 degrees Celsius cause
2 answers:
Answer:
False.
Explanation:
Hello,
In this case, as the moles are held constant, one uses the Gay-Lussac's law in order to relate both the temperature and the pressure at the beginning and at the end of the experiment via the following relationship:
By solving for the pressure at the second (final) state we can substantiate whether the 1.5 atm are correct as shown below:
Therefore, even when the pressure is correctly decreased, the given value is wrong as 2.8 atm is the correct final pressure, in this manner, the statement is false.
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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