Answer the following questions relating to HCl, CH3Cl, and CH3Br. — Intermolecular Forces Chemistry Question
Problem Context
Answer the following questions relating to HCl, CH3Cl, and CH3Br.
Model Answer
No, the student is not correct. None of the oxidation numbers of the elements change (H = +1, S = +6, O = -2, Na = +1, Cl = -1).
1 point is earned for indicating the student is incorrect and for providing a valid justification.
Model Answer
3.00 g HCl × (1 mol HCl / 36.46 g HCl) × (2 mol NaCl / 2 mol HCl) × (58.44 g NaCl / 1 mol NaCl) = 4.81 g NaCl
1 point is earned for determining the number of moles of HCl(g) (may be implicit).
1 point is earned for calculating a mass of NaCl based on the number of moles of HCl(g).
Model Answer
The pressure will remain the same. The reaction stoichiometry shows that two moles of gaseous reactants produce two moles of gaseous products. Because the number of moles of gas does not change, the pressure does not change.
1 point is earned for the correct answer and a valid justification.
Model Answer
The value of Kp is large, so the reaction will proceed to the right until the limiting reactant is essentially used up. Thus practically all of the CH3OH(g) will react and the final pressure of HCl(g) is 0.600 − 0.250 = 0.350 atm.
OR
ICE Table:
CH3OH(g) + HCl(g) ⇄ CH3Cl(g) + H2O(g)
I: 0.250 atm, 0.600 atm, 0, 0
C: -x, -x, +x, +x
E: 0.250 - x ≈ 0, 0.600 - x ≈ 0.350, +x ≈ 0.250, +x ≈ 0.250
The final pressure of HCl(g) is 0.350 atm at equilibrium.
1 point is earned for the correct answer.
Model Answer
Agree. The large value of Kp means that the partial pressure of the limiting reactant at equilibrium will be extremely small, but some CH3OH molecules must exist for the system to be in dynamic equilibrium.
OR
Kp = (p_CH3Cl * p_H2O) / (p_CH3OH * p_HCl) = 4.7 × 10^3 = (0.250 * 0.250) / (p_CH3OH * 0.350)
p_CH3OH = 3.8 × 10^−5 atm
The partial pressure of CH3OH(g) is very small but is not zero.
1 point is earned for the correct choice and a valid justification.
Model Answer
London dispersion forces and dipole-dipole forces.
1 point is earned for identifying both types of forces.
Model Answer
The electron cloud in CH3Br is larger and more polarizable than that of CH3Cl. As a result, the London dispersion forces are stronger in CH3Br compared to those in CH3Cl and consequently the boiling point of CH3Br is higher than that of CH3Cl.
1 point is earned for a valid explanation.
Model Answer
PV = nRT
n = 1.00 g CH3Cl / (50.5 g/mol CH3Cl) = 0.0198 mol
P = nRT / V = (0.0198 mol × 0.0821 L atm mol^−1 K^−1 × 298 K) / 0.00200 L = 242 atm
1 point is earned for the correct pressure with the correct units.
Model Answer
At room temperature the liquid will vaporize. Consequently the glass vial may not be strong enough to withstand the increase in pressure.
1 point is earned for a valid explanation.