<span> the </span>vapor pressure<span> of the liquid at a temperature T</span>2<span> ... Now, </span>it's<span> important to realize that the </span>normal boiling point<span> of a substance is measured at an atmoshperic ... ΔHvap=−ln(</span>134mmHg760mmHg<span> )⋅8.314J mol−1K−1 (1(273.15+</span>0)−1(273.15+40))K−1 ... Give equations that can be used tocalculate<span> the .
Now try it yourself :)</span>
.07! you divide 7 by 100%
Answer:
a) K = 5.3175
b) ΔG = 3.2694
Explanation:
a) ΔG° = - RT Ln K
∴ T = 25°C ≅ 298 K
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔG° = - 4.140 KJ/mol
⇒ Ln K = - ( ΔG° ) / RT
⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))
⇒ Ln K = 1.671
⇒ K = 5.3175
b) A → B
∴ T = 37°C = 310 K
∴ [A] = 1.6 M
∴ [B] = 0.45 M
∴ K = [B] / [A]
⇒ K = (0.45 M)/(1.6 M)
⇒ K = 0.28125
⇒ Ln K = - 1.2685
∴ ΔG = - RT Ln K
⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )
⇒ ΔG = 3.2694
