Respuesta :
Answer:
- -662 kJ/mol
Explanation:
The standard heat of formation is the enthalpy change for the formation of a compound from the elements at their standard states.
PCl₅ is formed by P and Cl, whose standard states are solid and diatomic gas.
You just must manipulate the two given equations to obtain an overall equation from P(s) and Cl₂(g).
Based of Hess' law, the enthalpy change of the overall equation is the algebraic sum of the individual equations.
If you multiply the first equation by 2 and reverse it you get:
[tex]2Cl_2(g)+2PCl_3(g)\rightarrow 2PCl_5(s);\Delta H^0rxn=-2mol\times 87.9kJ/mol[/tex]
Write the second equation:
[tex]2P(s)+3Cl_2)(g)\rightarrow 2PCl_3(g);\Delta H^0rxn=2mol\times (-574)kJ/mol[/tex]
Add them:
[tex]2P(s)+5Cl_2(g)\rightarrow 2PCl_5(s)[/tex]
[tex]\Delta H^0rxn=-2mol\times 87.9kJ/mol-2mol\times 574kJ/mol[/tex]
After computing:
[tex]\Delta H^0rxn=-1,323.8kJ[/tex]
Notice that the last calculation results in the standard enthalpy of formation for two moles of PCl₅(s). Thus, to obtain the enthalpy of formation per mol you must divide by 2:
[tex]\Delta H^0rxn=-661.9kJ/mol[/tex]
Round to 3 signficant figures: -662kJ/mol
Answer: -375 kJ/mol
Explanation:
First, write the equation for the formation of PCl5 starting with the elements P and Cl in their standard states. Note that the standard state of phosphorus is P(s) and the standard state of chlorine is Cl2(g).
P(s) + 5/2Cl2(g)⟶PCl5(s)
Now, determine how to the thermochemical equations above can be combined to yield this equation. If we reverse the direction of the first equation, and divide the second equation by two, the two equations can be combined as follows.
when you add the reactions:
PCl3(g) + Cl2(g) --> PCl5(s) Hrxn = -1 X 87.9 kJ/mol
and
P(s) + 3/2 Cl2(g) --> PCl3(g) Hrxn = 1/2 X -574 kJ/mol
you should get
P(s) + 5/2 Cl2(g) --> PCL5(s) Hrxn = Hf
To determine the enthalpy of formation for PCl5, combine the reaction enthalpies for two equations given as follows.
ΔH∘f = −1×87.9 kJ/mol + 1/2 × −574 kJ/mol
ΔH∘f = −374.9 kJ/mol
Rounding the answer to three significant figures, we find that the enthalpy of formation of PCl5(s) is approximately equal to −375kJmol.
Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.
