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Answer:
230.3 kJ
Some chemical reactions show release of heat while some show absorption of heat. On these bases, the reaction is classified as either endothermic or exothermic. Negative sign of ΔH means exothermic reaction.
Mass of water : 0.45 g
<h3>Further explanation</h3>
Given
1.5 g ethanoic acid
Required
Mass of water
Solution
Reaction
<em>CH₃COOH + NaOH ⇒H₂O + CH₃COONa</em>
mol CH₃COOH(MW=60 g/mol) :
= mass : MW
= 1.5 : 60
= 0.025
mol H₂O from the equation :
= 0.025(mol ratio CH₃COOH : H₂O= 1 : 1)
Mass of water :
= mol x MW water
= 0.025 x 18 g/mol
= 0.45 g
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Answer: Fe2O3
Explanation: By applying crisscross method, the formula will be Fe2O3.
