I think it's E. heating of water
Because exothermic process discharges heat, causing the temperature of the prompt environment to rise
Please correct me if I'm wrong!! I'd be happy to fix it!! :)
I think it would be C.100.5cm or D.100.5ml hope that helps
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.
A goes with u, C with G, and T with A
Reaction of methyl amine with HCl (H+) is:
CH3NH2 + H⁺ ↔ CH3NH3⁺
At equivalence point:
moles of CH3NH2 = moles of HCl
moles of CH3NH2 = 0.350 * 0.125 = 0.0438 moles = moles of HCl
Volume of HCl = 0.0438 / 0.250 = 0.1752 L
Therefore, total volume at eq point = 0.350 + 0.1752 = 0.5252 L
Now, at equivalence point there is 0.0438 moles of CH3NH3+. The new equilibrium is :-
CH3NH3+ ↔ CH3NH2 + H+
where the concentration of CH3NH3+ = 0.0438/0.5252 = 0.083 M
Ka = [CH3NH2][H+]/[CH3NH3+]
Kw/Kb = x²/0.083-x
10⁻¹⁴/4.4*10⁻⁴ = x²/0.083-x
therefore x = [H+] = 1.37*10⁻⁶M
pH = -log[H+] = -log[1.37*10⁻⁶] = 5.86
