1) What is the definition of the Enthalpy of Reaction? A) The amount of heat in

Question

1) What is the definition of the Enthalpy of Reaction?

A) The amount of heat in Joules require to evaporate 1 mole of reactant.

B) The amount of energy released from gaseous expansion during a reaction.

C)The amount of heat in kJ released (exothermic) or gained (andothermic) per mole of product.

2) Which of the following reactions needs to be reversed to solve for the Heat of Neutralization between Hydrochloric Acid and Ammonia?

A)Acetic Acid + Ammonia.

B)Hydrochloric Acid + Sodium Hydroxide.

C)Acetic Acid + Sodium Hydroxide.

3) What is the molar heat of neutralization (Enthalpy of Reaction) in kJ/mol if 5 moles of HA neutralized 5 moles of BOH and released 6578 J of heat?

4) Calculate the molar heat of neutralization in kJ/mol of the reaction between HA and BOH given the following information:

The temperature change equals 8C,
50 mL of 1 M concentration of Acid
50 mL of 1 M concentration of Base
Heat capacity of the calorimeter is 6.5 J/C.
Remember that the specific heat of water is 4.18 J/gC

5) How much heat is gained (in Joules) by the water where a chemical reaction takes place in 100 mL aqueous solution and has a temperature increase of 15 C?

Explanation / Answer

1] C

2] C

3] Molar heat (enthalpy) of neutralisation is the energy liberated per mole of water formed during a neutralisation reaction.

so Heat of neutralization : 6578 / 5 = 1.325 KJ/mol

4] V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL

V(HCl) = volume of HCl(aq) added to achieve neutralisation = 50.0 mL
c(NaOH) = concentration of NaOH(aq) = 1.0 mol L-1
c(HCl) = concentration of HCl(aq) = 1.0 mol L-1
  d = density of solutions = 1 g mL-1 (assumed)
Cg = specific heat capacity of solutions = 4.18 JoC-1g-1 ( assumed to water heat capacity )
q = heat liberated during neutralisation reaction = ? J

Check the units for consistency and convert if necessary:
Convert volume of solutions (mL) to mass (g): density x volume = mass
since density = 1 g mL-1: 1 x volume (mL) = mass (g)
mass(NaOH) = 50.0 g
mass(HCl) = 50.0 g

Calculate the heat produced during the neutralisation reaction:
heat produced = total mass x specific heat capacity x change in temperature
q = mtotal x Cg x T

delta T = 8C

mtotal = mass(NaOH) + mass(HCl) = 50.0 + 50.0 = 100.0 g
Cg = 4.18 JoC-1g-1
T = 8 C

q = 3344 J

Calculate the moles of water produced:
OH-(aq) + H+(aq) H2O(l)
1 mol OH-(aq) + 1 mol H+(aq) 1 mol H2O
moles(H2O) = moles(OH-(aq))
moles(OH-(aq)) = concentration (mol L-1) x volume (L) = 1.0 x 50.0/1000 = 0.050 mol
moles of water produced = 0.050 mol

Calculate the heat liberated per mole of water produced, Hneut :
Hneut will be negative because the reaction is exothermic
Hneut = heat liberated per mole of water = -1 x q ÷ moles of water
Hneut = -1 x 3344 ÷ 0.050 = -66.88KJ mol-1

5] density of water = 1gm/ml

100 ml -----> 100 *1 = 100 gms

q = mSdT

q = 100*4.18 *15 = 6270 J (heat gained by water )

all heat from the reaction comes from the reaction is taken by water

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