Hess Law

Questions

For A. through E. See the calculations for the Data Tables above.

A. Using the data from your data tables calculate ΔT for all three reactions:

Reaction 1 = Final – Initial =37.9-24.9=13.8 C

Reaction 2=Final –Initial=25.72-25.2= .52 C

Reaction 3=Final –Initial= 36.9-25.05=12 C

B. Calculate the heat loss or gain of the three solution mixtures:

Mass=Density*Volume=1.02*20=20.4 grams

Qsoln= Mass * Specific heat* Δ T

Reaction 1= 20.4g * 4.184 * 13= 1110 J

Reaction 2=20.4g * 4.184 * 0.52= 44.4 J

Reaction 3=20.4g * 4.184 * 12= 1024 J

C. Use Hess’ Law and ΔH for the first two reactions: NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq)

NaOH (aq) + NH4Cl (aq) → NH3 + NaCl + H2O (l)

to determine ΔH for this reaction: NH3 + HCl → NH4Cl

ΔHreaction 1 = -1100 / .02 mol = -55500 J = -55.5 kJ

ΔHreaction 2 = -44.04 / .02 mol = -2220 J = -2.22 kJ

HCl + NaOH → NaCl + H2O -55.5 KJ

NH3 + NaCl + H2O → NH4Cl + NaOH +2.22 KJ

NH3 + HCl → NH4Cl -52.35 KJ

D. Compare the results of step 3 above with the experimental results of the

NH3 + HCl → NH4Cl

ΔHreaction 1 = -1024 / .02 mol = -51200 J = -51.2 kJ

The difference between the 2 reactions calculated above and the Hess law is 1.15 KJ

E. Use the thermodynamic quantities given below to calculate the theoretical ΔH for this reaction: NH3 + HCl → NH4Cl

● ΔH°f for NH3 (aq) = - 80.29 kJ/mol

● ΔH°f for HCl (aq) = - 167.2 kJ/mol

● ΔH°f for NH4 (aq) = - 132.5 kJ/mol

● ΔH°f for Cl- (aq) = - 167.2 kJ/mol

ΔH = ∑ ΔH°f (products) - ∑ ΔH°f (reactants)

ΔH = (-132.5 + -167.2) - (-80.29 + -167.2)

ΔH = -52.21 KJ

F. What was the ΔH value obtained for NH3 + HCl→ NH4Cl from Hess’ Law method?

ΔH = -52.35 KJ (From question C above)

G. What was the ΔH value obtained for NH3 + HCl →NH4Cl experimentally?

ΔH = -51.2 KJ (Question D above)

H. What was the calculated ΔH value...