Calculate the heat evolved when 75 0 mL of a 2.00M HCl solution reacts with 37.5
ID: 950486 • Letter: C
Question
Calculate the heat evolved when 75 0 mL of a 2.00M HCl solution reacts with 37.5 mL or a 4 00M NaOH solution if the temperature rises from 20 0 degree C to 37.9 degree C. (Remember you are calculating this by an indirect measurement of heat absorbed by the solution.) How many moles of water are produced by the actual amount of reactants in Problem #1? What is the molar heat of neutralization in the Problem #1? What is the net ionic equation for the reaction in Problem #1? What are the spectator ions? Based on your answer to Problem #3. what would be the heat of reaction for the b net ionic equation: 3OH^- (aq) + 3H^+ (aq) rightarrow 3H_2 O(I)Explanation / Answer
Let's assume;
1ml = 1gr for laboratory units!
Total volume of reactants = 0.075 L + 0.0375 L = 0.1125 L
If Total mass of reactant = 0.1125 kg = 112.5 gr (due to asumption)
T = 37.9°C - 20°C = 17.9°C
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Heat of Neutralization
H = - Q/n (KJ/mol)
1.Heat evolved:
Q = mcT
c = specific heat of water 4.2 J/g.°C
Q = 112.5gr x 4.2 J/g.°C x 17.9°C = 8457.75 J
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NaOH + HCl ----> NaCl + H2O
from here you can see this is a reaction between acid and base
H+ + OH- ----> H2O
M = mol/ L
mol H+ = 2 mol/L * 0.075 L = 0.15 moles
mol OH- = 4 mol/L * 0.0375 L = 0.15 moles
2.Moles of water
mol H2O = 0.1125 gr * 0.15 mol = 0.0168 mol
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3.Heat of neutralization
H = 8457.75 J/0.0168 mol = 501.20 KJ/mol
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4.Net ionic equation
H+Cl- + Na+OH- ----> Na+ Cl- + H+ OH-
H+ + OH- ----> H2O reaction, so Na+ Cl- are spectators.
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5.Heat evolved of new Reaction
3H+ + 3OH- ----> 3H2O
#moles = 3
H = - Q/n (KJ/mol)
500.20 KJ/mol = Q/3(0.0168) moles
Q = 25.21 KJ
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