Heat sublimation ? not in my book or anywhere online
The standard heat of formation of PI3 (s) is -24.7 and the bond energy in this PI molecule is 184 . The standard heat of formation of P (g) is 334 and that of I2 (g) is 62 . The I2 bond energy is 151 .
Calculate the heat of sublimation of .
PI3 [PI3(s) –> PI3 (g) ]
not even in my book or not talked about in class but somehow is given as a hw problem go figure
This is the best answer for Heat sublimation?
_____I answered this Q before. You said I was wrong; maybe you can tell me why. Did you use the sum of the eqns? I did and they do give the right answer. If you use the ~value I supplied, that was about right, then you got the wrong answer; I expected you to do the math. I also gave a ref of what the answer should be; (the sum of the dHs just might NOT give that number within 1%!
to 1st approximation: heat sublimation of PI3 ~ (MW PI3 / MW I2) * heat sublimation of I2.
I have given the answer below, again.
_____Bond energy and enthalpy combination and HESS
_____1/4P4(s)—————–> P(g)_______dHvap = 334 kJ/mole P
___3/2 * [I2(s) ——————> I2 (g)]_____dHvap = 3/2 * 62 kJ/mole I2
_____P(g) +3/2 I2(g)———> PI3(g)____dHrxn= – [(3 bonds*184) – 3/2 bonds*151]kJ/mole
sum eqns and dHs above
_____1/4P4(s) + 3/2 I2(s) —> PI3(g)_____dHf PI3(g) = sum dHs = ?? kj/mole
_____PI3(s) ——–> 1/4P4(s) + 3/2 I2(s)__-dHf PI3(s) = 24.7 kJ/mole
sum eqns and dHs for these two reactions
_____PI3(s) ——————–> PI3 (g) ___________dHsubl = ??
SOLVE (answer should be ~ 95 kJ/mole; not ~400 kJ/mole)
Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.
22 hours ago
0.98 eV heat of sublimation of PI3 = 94.56 kJ/mole
1 eV = 96.48530891 kJ / mole (wikipedia)
Asker’s Comment: not right but thanks for answering
Start with PI3(s).
Take it apart into the component elements
vaporize those elements
recombine them to PI3(g)
The total energy change of this will be the heat of sublimation.
I think this is the correct answer for what is Heat sublimation?
if “the bond ability in this PI molecule is 184”, for in line with mole of P_F bonds. then the reaction is 3X’s 184kJ: (A) PI3 (g) –> P(g) & 3 I(g)….. (+552kJ) “The I2 bond ability is 151 ” m for the reaction: (B) I2 (g) –> 2 I (g) ….. (+151kJ) “the huge-unfold warmth of formation of PI3 (s) is -24.7”, for the reaction: (C) P(s) & a million.5 I2(s) –> PI3 (s) ….(-24.7kJ) “the huge-unfold warmth of formation of P (g) is 334”, for the reaction: (D) P(s) –> P(g) …. (+334 kJ) “the huge-unfold warmth of formation of I2 (g) is sixty two”, for the reaction: (E) I2 (s) –> I2 (g) … (+sixty two kJ) ==================== Calculate the warmth of sublimation of : PI3(s) –> PI3 (g) could be stumbled on by utilising, the opposite of equation (C), to supply PI3 (s) as a beginning cloth the opposite of equation (A), to supply PI3 (g) as a product -(C) : PI3 (s) –> P(s) & a million.5I2(s) ….(+24.7kJ) -(A) : P(g) & 3 I(g) –> PI3 (g) ….. (-552kJ) then to calcel out undesireables: +(D): P(s) –> P(g) …. (+334 kJ) cancels out P(s) & P(g) +a million.5(B)’s: a million.5 I2 (g) –> 3 I (g) ….. (a million.5)(+151kJ) cancels out I(g) & I2(g) ============== hess’s regulation says that if combining equations -A , -C , +D , & a million.5B can supply the equation which you go with… then combining the flexibility from -A , -C , +D , & a million.5B can supply the flexibility which you go with… (-dHA of +24.7) & ( -dHC of -552) , +dHD of 334) , & (a million.5dHB of 151 = 226.5) = your answer: 33.2 kJ you would be able to desire to contemplate rounding off for sig figs to dH = 33 kJ, through fact a lot of your dH’s weren’t measured to the 0.1th of a kJ