Given δh°f for so2 (g ) (-296.8 kj/mol) and so3 (g ) (-395.7 kj/mol), calculate δh°rxn for the conversion of so2 to so3.

Hey there!:

Given that;

SO2 (g ) = (-296.8 kJ/mol) and SO3 (g )= (-395.7 Kj/mol)

2 SO2(g) + O2(g) > 2 SO3(g)

ΔH reaction = ΔH of product - ΔH of reactant

We know that  ΔH O2 = 0 kj

ΔH reaction = [2 x - 395.7] - [2 x - 296.8 +0.0 KJ]

= - 791.4 - (- 593.6)

= - 791.4 + 593.6

= - 197.8 kJ

Because there are 2 mole of SO2 coverted into 2 mole of SO3 :

=- 197.8 kJ/2 mol

ΔH°rxn  = - 98.9 kJ/ mol

Hope that helps!

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.