Question

Using the Table of Bond Energies in CH 11.5 calculate the ΔH for : H₂(g) + F₂(g) → 2HF(g)

assume: H-H ; F-F ; H-F bonds

	A.	-543
	B.	-24
	C.	24

Given: A(g) + B(g) → D(g) ; ΔH = 56kJ/mol ; the enthalpy change for 2A(g) + 2B(g) → 2D(g) is:

	A.	-56
	B.	-112
	C.	112

Given: A(g) + B(g) → C(g) ΔH = 10kJ/mol ; C(g) → D(g) ΔH = 30kJ/mol;

the ethalpy change for: A(g) + B(g) → D(g) is _____ kJ/mol

	A.	20
	B.	-20
	C.	40

Answer 1

we can solve this type of problems using HESS LAW .

HESS LAW OF CONSTANT HEAT SUMMATION :-

It states that regardless of different stages or steps of the reaction,the total enthalpy change (∆H) of the reaction is sum of all changes. This is because enthalpy is s state function.

Also if any mathematical operation is done to the reaction same should be done with the value of enthalpy.

In the first question you have not given the value of bond energies, so i have taken standard data from wikkipedia

So my Answer is different from your option. Although you can apply same method with your to get your option

But it quite close to option A. So the answer will be A

(solurion) ( SORUSH fortH, = 0 kj/mol. DH for F = 0 kj/mol oH for HF = -273.3 Kylmat. Given He (9) + F2 (9) L2HF (9) So, by Hesslow, - reaction - Product = 2x D ) - ( 1x ans + Julier, = 2X(-273-3) - (4x0 +1vo) TDH =.-546 6. Kij/mot | Ans: Given, A (g) + BCA) - D (g) DH = 56 kylned - 0 Now, DH for, 2A (9) + 2B1g) — 20 (9) - @quation © = 2x @pewrim ] { from Hess low> So, DH fou ☺ = 2 * SH for D 1. .= 2 x 56, SDH = 112 kj/mol Ang So option c is correst

e Giren, A(g) + B (g) Cigo DH, = 10 kJ hmot C (g) + DL9) DH= 30 kJ/mol. Now, adding above two reaction. A (9) +B (g): =.cglish, . cigo - Digi Dha A (g) + B (g) = D(g) DH so, by Hess law, if we add two aton its enthaly will also be added DH = DH, TPH2 = 10 +30 (on s 40 kg /mol Ans option c is coobert

Don't forget to upvote my Answer!!