The following question was sent by one of my A-Level JC1 H2 Chemistry students in Singapore.
It is a very interesting MCQ question and requires the students to be very good in drawing the Energy Cycle Diagram, in order to solve the question.
She thinks that the answer should be “B”‘. Do you agree with her?
Show your working or brief explanation on arriving at your answer.
I will share with everyone on the solution soon.
“80% of the Success is about taking the Necessary Actions”
yeah i think its B too. release the soln pls!!!
The enthalpy of formation is represented by the equation:
1/2I2 + 3/2Cl2 –> ICl3
To get this you can multiply equation 1 by 1/2 and then add equation 2
Hf(ICl3) = +7 – 88 = -81 kJ
My answer is the same as hers. (+14/2)+(-88)=-81kJ/mol.
I2 + Cl2 (+2Cl2) —————–> 2ICl3 {1}
2ΔHf(ICl3)
I2 + Cl2 (+2Cl2) —————> 2ICl + 2Cl2 {2}
+14 kJmol^-1
2ICl +2Cl2 ——————–> 2ICl3 {3}
2(-88) kJmol^-1
It’s difficult to show the trianglular diagram here, so just try to visualize the triangle. Ignored the state symbols to avoid getting messy. Won’t be doing that in the real exams though.
In my working, I’d put eqn {1} on top, and {2} and {3} in the left and right bottom corners of the diagram respectively.
You can see that the (+2Cl2) in {1}will sort of “balance” eachother out, so there’s no need to consider bond energy of Cl2.
By Hess’ Law, 2ΔHf(ICl3)= 14+2(-88)= -162 kJmol^-1
∴ ΔHf(ICl3)= -81 kJmol^-1
So I guess the girl would have been correct, answer B.
Let me know if you spot any errors, thanks.
Yes,I’m agree with her asnwer~
Firstly, we will get an answer (-162 kJ per mole) for 2 iodine trichloride~ since enthalpy change of formation is meant for 1 mole , we have to divide -162/2 and thus we will get -81 kJ mol-1 as our answer~ Is my concept correct, Sir? thanks for guiding me~