Answer:
ΔrxnH = -580.5 kJ
Explanation:
To solve this question we are going to help ourselves with Hess´s law.
Basically the strategy here is to work in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.
Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-
The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.
This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.
Taking 1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.
Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l) ΔrxnHº = 259 / 2 kJ 1/2 eq3
+ 2NaOH(s) + SO₃(g) ⇒ Na₂SO₄ (s) + H₂O (l) ΔrxnHº = -418 kJ - eq 2
+ 2Na (s) + 2 H₂O (l) ⇒ 2 NaOH (s) + H₂ (g) ΔrxnHº = -146 x 2 2 eq 1
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Na₂O (s) + SO₃ (g) ⇒ Na₂SO₄ (s) ΔrxnHº = 259/2 + (-418) + (-146) x 2 kJ
ΔrxnH = -580.5 kJ
The reaction N2O4 (g) <--> 2NO2 (g) is endothermic, meaning that it consumes heat to move towards formation of the products.
According to Le Chatelier's Principle, therefore, if heat is added, more product (NO2) will be produced, and equilibrium would shift towards the right side. This is choice 3.
