Thermodynamics & Solubility Experiment: KNO3(s)+H2O(l)--->K^1(aq)+NO3^-1(aq) 1)

Question

Thermodynamics & Solubility Experiment: KNO3(s)+H2O(l)--->K^1(aq)+NO3^-1(aq)

1) What is the value of the equilibrium constant in this experiment when 13.012 grams of KNO3 dissolves 13.70 ml of water?

2) What are the units of the slope of a line from a graph of lnK vs 1/T, where T is measured in Kelvin units? Show your work.

3) A linear regression analysis for this lab is y=-18.5x+5.96. Calculate the value of the enthalpy and entropy of the reaction. THEN, use your results to calculate the standard Gibbs Energy value at 298 K. Watch your units.

Explanation / Answer

Q1.

Ksp = [K+][NO3-]

calculate molarity o KNO3

mol of KNO3 = mass/MW = 13.012/101.1032 = 0.1287001 mol

V = 13.70*10^-3 L

M = mol/V = 0.1287001 / (13.70*10^-3) = 9.39 M

so

Ksp = 9.39*9.39

Ksp 88.1721 for KNO3

Q2.

units of slope whne we use

lnK vs 1/T

then

must be those of T

since lnK is dimensionless and

1/T must be cancelled so units must be those of T (KELVIN)

Q3.

enthalpy and entropy of reaction

slope = -dhvap/R

-18.5 = -dHVap/8.314

Hvap = (18.5*8.314) = 153.809 J/molK

entropy

intercept = dS/R

dS = R*intercept = 5.96*8.314 = 49.55144 J/molK

Calculate dG

dG = dH - T*dS = (153.809) - 298*49.55144 = -14612.520 J/mol

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