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

Answer:

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.

Step-by-step explanation:

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

$Molarity=\frac{Moles}{Volume (L)}$

Molarity of HCl = 0.250 M

Volume of HCl = 100 ml = 0.1 L

Moles of HCl = n

$n=0.250\times 0.1 L=0.0250 mol$

Molarity of NaOH= 0.150 M

Volume of NaOH= 200 ml = 0.2 L

Moles of NaOH= n'

$n'=0.150\times 0.2 L=0.030 mol$

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.

$\frac{1}{1}\times 0.0250 mol=0.0250 mol$ 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:

$Q=\Delta H^o\times 0.0250 mol=-58 kJ/mol\times 0.0250 mol=-1.45 kJ$

(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

$m=d\times v=1.00 g/mL\times 100 mL=100 g$

Mass of NaOH solution = m'

Volume of NaOH ,v' = 200 ml

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

$m'=d'\times v'=1.00 g/mL\times 200 mL=200 g$

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 = ?

$1450 J=300g\times 4.19 J/g^oC\times \Delta T$

$\Delta T=\frac{1450 J}{300g\times 4.19 J/g^oC}$

$\Delta T=1.15^oC$

1.15°C is the temperature increase.