Question

A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The temperature of both solutions and the calorimeter was
25.2°C before mixing and 31.3°C after mixing. The heat capacity of the calorimeter is 24.2 J/K.
Calculate the ΔHrxn for this reaction in units of kJ / mol of copper (II) hydroxide (19 points). Assume
the solutions is dilute enough that the specific heat and density of the solution is the same as that
of water (

Answer 1

Answer:

ΔH/mol Cu(OH)₂= 34685.95 J/mol H₂O =34.68595 kJ/mol H₂O

Explanation:

The reaction that occurs in this case is:

CuSO₄ + 2 KOH → Cu(OH)₂ + K₂SO₄

The measurement and calculation of the amounts of heat exchanged by a system is called calorimetry. The equation that allows calculating these exchanges is:

Q=c*m*ΔT

where Q is the heat exchanged for a body of mass m, constituted by a substance whose specific heat is c, and ΔT is the temperature variation experienced.

In this case:

• c=1 $\frac{cal}{g*C}$=4.184  $\frac{J}{g*C}$ because the specific heat and density of the solution is the same as that of water
• Assuming that the total volume is the sum of the individual volumes then:

total volume= volume of CuSO₄ + volume of KOH = 150 mL + 150 mL

total volume= 300 mL

Given that the solution has the same density as water (1.00 $\frac{g}{mL}$) then

$mass solution=300 mL*\frac{1 g}{1 mL}$

mass solution = 300 g

• ΔT=31.3° C - 25.02 °C= 6.1° C

Then

Q = 300 g* 4.184  $\frac{J}{g*C}$*6.1° C =7656.72 J

On the other hand, the heat absorbed by the calorimeter is ( taking into account that to convert from degrees C to degrees K, 273.15 must be added):

24.2 $\frac{J}{K}$ * [(31.3+273.15) - (25.2+273.15)] K = 147.62 J

So,  the heat exchanged throughout the system is

Qsystem=7656.72 J + 147.62 J=7804.34 J

The reactants are present in their stoichiometric ratio, so there is no excess.

By reaction stochetry (that is, by the relationships between the molecules or elements that make up the reactants and reaction products) 1 mol of CuSO₄ produce 1 mol of Cu(OH)₂.

Knowing that moles of CuSO₄ are:

$molesCuSO4=150 mL*\frac{1 L}{1000 mL}*\frac{1.50 mol}{1 L}$

moles CuSO₄=0.225

You can apply the rule of three.

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So

$x=\frac{c*b}{a}$

In this case:  If 1 moles of Cu(OH)₂ are formed if 1 moles of CuSO₄ react, how many moles of Cu(OH)₂ will be formed if 0.049 moles of CuSO₄ react?

$moles Cu(OH)2=\frac{0.225moles*1mol}{1 mol}$

moles Cu(OH)₂=0.225 moles

Now

ΔH/mol Cu(OH)₂ = $\frac{Qsystem}{moles Cu(OH)2} =\frac{7804.34 J}{0.225 moles}$

ΔH/mol Cu(OH)₂= 34685.95 J/mol H₂O =34.68595 kJ/mol H₂O

The conversion was done remembering that 1 kJ = 1000 J

Answer 2

Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
mols KOH = 3.00 x 0.150 = 0.450 
specific heat solns = specific heat H2O = 4.18 J/K*C 

CuSO4 + 2KOH = Cu(OH)2 + 2H2O 
q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T 
q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) 
dHrxn in J/mol= q/0.225 mol CuSO4 
Then convert to kJ/mol