Answer:
a) NaOH is the limiting reactant
b) the heat capacity of the system = 418.4 J/K
c) This is an exothermic reaction. The system will release energy on the surroundingd. The surroundings will absorb energy. (502.08J)
d) The reaction releases 502.08 J of heat
e) the molar enthalpy = 55.8 kJ/mol
Explanation:
Step 1: Data given
50.0 mL of 0.180 M sodium hydroxide solution is added to 50 mL of 0.200 M lactic acid solution.
CH3CHOHCO2H + OH- → CH3CHOHCO2- +H20
Step 3: Calculate moles of NaOH
Moles NaOH = Molarity of NaOH * volume
Moles NaOH = 0.180 M * 50*10^-3
Moles NaOH = 0.009 moles
Step 4: Calculate moles of lactic acid
Moles lactic acid = 0.2M *50*10^-3 L
Moles lactic acid = 0.01 moles
Step 5: Calculate limiting reactant
For 1 mole lactic acid, we need 1 mole of NaOH (mole ratio is 1:1)
NaOH has the smallest number of moles. This means <u>NaOH is the limiting reactant.</u> It will completely be consumed (0.009 moles)
There will be remain 0.01 - 0.009 = 0.001 mole of lactic acid.
<em>b. The system can be regarded as 100 g of water, C= 4.184 J/K°mol Calculate the heat capacity of the system.</em>
q = 4.184 J/K*g * 100 grams
q = 418.4 J/K
<em>c) The initial temperature of the system is 22.47 deg. Celsius; the final temperature is 23.67 deg. Celsius. Calculate the amount of heat absorbed by the system.</em>
Since we have an increase of temperature this is an exothermic reaction. This means the system releases heat to the surroundings. The systems does NOT absorb heat. (0J)
The surroundings absorbs heat from the system (502.08 J)
Since they absorbe energy. The enthalpy ΔH will be positive here.
<em>d) Calculate the amount of heat released by the reaction</em>
q = 418.4 * ΔT
q = 418.4 *(23.67-22.47)
q = 418.4 * 1.2
q= 502.08 J
Since the heat is released ΔH (enthalpy) will be negative.
<em>e) Calculate the molar enthalpy of neutralization for lactic acid.</em>
ΔH =(502.08 J) / (0.009 moles = 55786.7 J/mol = 55.8 kJ/mol
Enthalpy will be negative, the molar enthalpy is positive.
