Question

How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed? HCl(????????)+NaOH(????????)⟶NaCl(????????)+H2O(????)ΔH°=−58kJ If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation?

Answer 1

1.45 kilo joules of heat is produced when 100 mL of 0.250 M HCl and 200 mL of 0.150 M NaOH.

1.15°C is the temperature increase.

Explanation:

`HCl+NaOH\rightarrow NaCl+H_2O` ,ΔH°=-58 kJ/mol

`Molarity=(Moles)/(Volume (L))`

Molarity of HCl = 0.250 M

Volume of HCl = 100 ml = 0.1 L

Moles of HCl = n

`n=0.250* 0.1 L=0.0250 mol`

Molarity of NaOH= 0.150 M

Volume of NaOH= 200 ml = 0.2 L

Moles of NaOH= n'

`n'=0.150* 0.2 L=0.030 mol`

According to reaction, 1 mol of HCl reacts with 1 mol of NaOH. Then 0.0250 mole of HCl will reacts with 0.0250 mol of NaOH.

`(1)/(1)* 0.0250 mol=0.0250 mol` of NaOH

As we can see that moles of NaOH are in excess.Hence, excessive agent.

The enthalpy of the reaction = ΔH°=-58 kJ/mol

Energy released when 0.0250 moles of HCl reacted with 0.0250 moles of NaOH:

`Q=\Delta H^o* 0.0250 mol=-58 kJ/mol* 0.0250 mol=-1.45 kJ`

(Negative sign indicates that heat is liberated.)

Mass of the HCL solution = m

Volume of HCl ,v= 100 ml

Density of HCl solution = d = 1.00 g/mL

`m=d* v=1.00 g/mL* 100 mL=100 g`

Mass of NaOH solution = m'

Volume of NaOH ,v' = 200 ml

Density of NaOH solution = d' = 1.00 g/mL

`m'=d'* v'=1.00 g/mL* 200 mL=200 g`

Mass of the solution after mixing,M = m + m' = 100 g + 200 g = 300 g

Heat absorbed by the final solution formed after mixing = Q'

Heat absorbed by the final solution formed after mixing = Heat released during reaction

Q' = -Q = -(-1.45 kJ)= 1.45 kJ=1450 J (1kJ = 1000 J)

Q' = mcΔT

generally, m = mass of the substance

c = specific heat of the substance

ΔT = Change in temperature

Specific heat capacity of the product formed after mixing = c = 4.19 J/g°C

Mass of the resulting mix = 300 g

ΔT = ?

`1450 J=300g* 4.19 J/g^oC* \Delta T`

`\Delta T=(1450 J)/(300g* 4.19 J/g^oC)`

`\Delta T=1.15^oC`

1.15°C is the temperature increase.