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Anastaziya [24]
3 years ago
10

When 50.0 mL of 1.27 M of HCl(aq) is combined with 50.0 mL of 1.32 M of NaOH(aq) in a coffee-cup calorimeter, the temperature of

the solution increases by 8.49°C. What is the change in enthalpy for this balanced reaction? HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Assume that the solution density is 1.00 g/mL and the specific heat capacity of the solution is 4.18 J/g⋅°C. Hint: You need to determine the limiting reagent.
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

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>

<em />

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>

<em />

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