Answer:
-55.9kJ/mol is the change in enthalpy of the reaction
Explanation:
In the reaction:
HCl(aq) + NaOH(aq) → H₂O(l) + NaCl
Some heat is released per mole of reaction.
To know how many moles reacts we need to find limiting reactant:
Moles HCl = 0.050L ₓ (1.27mol / L) = 0.0635 moles HCl
Moles NaOH = 0.050L ₓ (1.32mol / L) = 0.066 moles NaOH
As there are more moles of NaOH than moles of HCl, <em>HCl is limiting reactant and moles of reaction are moles of limiting reactant, </em><em>0.0635 moles</em>
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Using the coffee-cup calorimeter equation we can find how many heat was released thus:
Q = C×m×ΔT
<em>Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)</em>
Replacing:
Q = 4.18J/g°C×100g×8.49°C
Q = 3548.8J of heat are released in the reaction
Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:
ΔH = Heat / mol of reaction
ΔH = -3548.8J / 0.0635 moles of reaction
<em>Negative because is released heat. </em>
ΔH = -55887J / mol
ΔH =
<h3>-55.9kJ/mol is the change in enthalpy of the reaction</h3>
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