Question

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ per mol of HCl, assuming that the calorimeter loses only a negligible quantity of heat. The total volume of the solution is 100 mL, its density is 1.0 g/mL, and its specific heat is 4.18 J/g*K.

Answer 1

54.34 kJ/mol of HCl

Step-by-step explanation:

The calorimeter is a device used to determine the heat that is lost or gained, by a reaction. When the temperature change without a phase change, the heat (Q) can be calculated by:

Q = m*c*ΔT

Where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the temperature variation (final - initial). The mass of the solution is the density multiplied by the volume:

m = 1.0 g/mL * 100 mL = 100 g

The temperature variation in °C is equal to the temperature variation in K, thus:

Q = 100g * 4.18J/gK * (27.5 - 21.0)K

Q = 2717 J

Thus, the solution gained 2717 J of heat. The enthalpy is how much of this energy is inside the matter, thus, it is the heat divided by the number of moles of a substance.

The number of moles of HCl is the volume (50 mL = 0.05 L) multiplied by the concentration:

n = 0.05 L * 1.0 M = 0.05 mol

The enthalpy is the heat divided by the number of moles:

H = 2717/0.05

H = 54340 J/mol of HCl

H = 54.34 kJ/mol of HCl

Answer 2

Answer: 54.4 kJ/mol

Step-by-step explanation:

First we have to calculate the moles of HCl and NaOH.

`\text{Moles of HCl}=\text{Concentration of HCl}* \text{Volume of solution}=1.0M* 0.05=0.05mole`

`\text{Moles of NaOH}=\text{Concentration of NaOH}* \text{Volume of solution}=1.0* 0.05L=0.05mole`

The balanced chemical reaction will be,

`HCl+NaOH\rightarrow NaCl+H_2O`

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water =
`50ml+50ml=100ml`

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1g/ml* 100ml=100g`

Now we have to calculate the heat absorbed during the reaction.

`q=m* c* (T_(final)-T_(initial))`

where,

q = heat absorbed = ?

`c` = specific heat of water =
`4.18J/g^oC`

m = mass of water = 100 g

`T_(final)` = final temperature of water =
`27.5^0C`

`T_(initial)` = initial temperature of metal =
`21.0^0C`

Now put all the given values in the above formula, we get:

`q=100g* 4.18J/g^oC* (27.5-21.0)^0C`

`q=2719.6J=2.72kJ`

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of
`HCl`:

0.05 moles of
`HCl` releases heat = 2.72 KJ

1 mole of
`HCl` releases heat =
`(2.72)/(0.05)* 1=54.4KJ`

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ