According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
Answer:
Reaction A and B are unfavorable.
Explanation:
Gibbs free energy is an energy which that can be use to convert into useful work.
ΔG°=ΔH°-TΔS°
ΔG°= Gibbs free energy
ΔH° = enthalpy of reaction
T = temperature of eh reaction
ΔS° = Entropy change
- If the Gibbs free energy of the reaction is positive than the reaction will be non spontaneous and the chemical reaction will be not feasible.
- If the Gibbs free energy of the reaction is negative than the reaction will be spontaneous and the chemical reaction will be feasible .
According to given information in the question:
Reaction A and B are non spontaneous as their Gibbs free energy value is positive.hence both are unfavorable.
Answer: the percent composition of carbon in heptane is 83.9%
Explanation:
<u>1) Atomic masses of the atoms:</u>
<u>2) Molar mass of heptane:</u>
- C₇H₁₆: 7 × 12.01 g/mol + 16×1.008 g/mol = 100.2 g/mol
<u>3) Mass of carbon in one mole of heptane:</u>
- C₇: 7 × 12.01 g/mol = 84.07 g/mol
<u>3) Percent composition of carbon:</u>
- % = (mass in grams of C) / (mass in grams of C₇H₁₆) × 100 =
= (84.07 g/ 100.2 g) × 100 = 83.9% ← answer
