Question

1:

A piece of solid magnesium is reacted with hydrochloric acid to form hydrogen gas:

Mg_(s) + 2 HCl_(aq) à MgCl_2(aq) + H_2(g)

What volume of hydrogen gas (in mL) is collected over water at 25 °C by reaction of 0.450 g of Mg (AW = 24.3 g/mol) with 5.00 mL of 1.0 M HCl? The barometer records an atmospheric pressure of 762 torr and the vapor pressure of water at this temperature is 23.35 torr.

2:

Given the following balanced chemical reactions and their enthalpies, calculate the enthalpy of reaction for 2C(s) + H₂(g) ---> C₂H₂(g):

C2H2(g) + (5/2) O2(g) ---> 2 CO2(g) + H2O(ℓ)	ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g)	ΔH° = -393.5 kJ
H2(g) + (1/2) O2(g) ---> H2O(ℓ)	ΔH° = -285.8 kJ

3:

A 10.00 mL aliquot of 0.100 M NaOH is added to 15.00 mL of 0.100 M HCl (don't try this at home, since acids and bases react violently to produce water and heat!) in a coffee cup calorimeter. The initial temperature of the two dilute liquids were both 25.0 ºC and the final temperature of the mixture was found to be     37.8 ºC. What is the (molar) heat of reaction for the neutralization?

4:

The heat of vaporization for toluene is 38.06 kJ/mol and the normal boiling point is 111 ºC. Using the Clausius-Clapeyron equation, calculate the vapor pressure of toluene at room temperature (25 ºC).

5:

What is the equilibrium constant for a weak acid HA that dissociates 1% at an initial concentration of 0.05 M according to the following balanced chemical equation?

HA_(aq) ↔ H⁺_(aq) + A^-_(aq)

Answer 1

1.

Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g)

No of mol of Mg = 0.45/24 = 0.01875 mol

No of mol of HCl = 5/1000*1 = 0.005 mol

limiting reactant is HCl

No of mol of H2 produced = 0.005/2 = 0.0025 mol

Volume of H2 = nRT/P

     = 0.0025*0.0821*298/((762-23.35)/760)

   = 0.063 L

    = 63 ml

2.

2 CO2(g) + H2O(ℓ) ----> C2H2(g) + (5/2) O2(g)   DHrxn = 1299.5 kj

2* ( C(s) + O2(g) ---> CO2(g) )               DHrxn = -393.5*2

      H2(g) + (1/2) O2(g) ---> H2O(ℓ)           DHrxn = -285.8

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   2C(s) + H2(g) ---> C2H2(g)   DHrxn = 1299.5+2*-393.5+(-285.8) = 226.7 kj

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