What is δs° for the reaction so2(s) no2(g) → so3(g) no(g)? substance s°(j/k • mol) so2(g) 248.5 so3(g) 256.2 no(g) 210.6 no2(g)
1 answer:
The value of ΔS° for reaction is - 22.2 J/K.mol
→
Calculation,
Given value of S°(J/K.mol) for
= 248.5
= 240.5
= 210.6
= 256.2
Formula used:
ΔS° (Reaction) = ∑S°(Product) - ∑S°(Reactant)
ΔS° = (256.2 + 210.6 ) - ( 248.5 + 240.5) = 466.8 - 489 = - 22.2 J/K.mol
The change in stander entropy of reaction is - 22.2 J/K.mol. The negative sign indicates the that entropy of reaction is decreases when reactant converted into product.
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(a) 33.6 L of oxygen would be produced.
(b) 106 grams of would be needed
1 mole of gas = 22.4 L
(a) From the equation, 2 moles of produce 3 moles of
. 1 mole of
will, therefore, produce 1.5 moles of
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1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L
(b) 22.4 L of is produced at STP. This means that 1 mole of the gas is produced.
From the equation, 1 mole of requires 1 mole of
.
Molar mass of = (23x2)+ (12)+(16x3) = 106 g/mol
Mass of 1 mole = 1 x 106 = 106 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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