10 Thermodynamics Pre-lab Quiz Exit Exam Question 8 of 10 Use the following information to answer questions 5-10. A student determined the heat of neutralization of copper (II) sulfate (CuSO4) mixed with potassium hydroxide (KOH) using the procedure described part 2 of this experiment. A 150.0 mL sample of a 1.50 M solution of CuSO4 was 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 heat of reaction (ΔHrxn) for this reaction in units of J/mole of copper (II) hydroxide (Cu(OH)2). Assume the solution is dilute enough that the specific heat capacity and density of the solution is the same as that of water, 4.18 J/g∙K and 1.00 g/mL respectively. Calculate the number of moles of copper (II) hydroxide (Cu(OH)2) formed during this reaction. Assume the reaction goes to 100% completion. 0.112 moles 0.150 moles 0.225 moles 0.45 moles Next

ΔHrxn = - 34,653 J/mol Cu(OH)₂

0.225 mol Cu(OH)₂

Explanation:

The reaction in this question is

CuSO₄   +   2  KOH   ⇒   Cu(OH)₂   +   K₂SO₄  ( double displacement )

Now notice we are told the temperature in both the calorimeter and solution increases from 25.2 ºC to 31.3 ºC after mixing. Thus we have an exothermic reaction and to calculate its heat of reaction we need to compute the heat absorbed by the calorimeter and the  solution  keeping in mind that the solution is dilute enough to assume the heat capacity of the solution and its density are the same as that of water.

q solution = m x c x ΔT

where m is the mass of solution, c is the specific heat of water, and ΔT is the change in temperature for the reaction.

We know the volume of mixing and the density of water so we can calculate m:

vol mix =  150.0 mL + 150.0 mL  = 300.0 mL

m = 300.0 mL x 1 g/ mL

T₀ = 25.2 ºC + 273 = 298.2 K

T₁ = 31.3 ºC + 273 = 304.3 K

ΔT = 304.3 K - 297.2 K = 6.1 K

q solution  = 300.0 g x 4.18 J / g·K x  6.1  K =  7649.4 J

q calorimeter = C calorimeter x ΔT = 24.2 J/K x 6.1 K = 147.6 K

q total = 7649.4 J + 147.6 K = 7797 J

Now we have the total heat of the reaction but we are asked the change in enthalpy per mole of Cu(OH)₂.

To determine the mol Cu(OH)₂ formed we need to perform a calculation based on the stoichiometry of the reaction.

To do that first calculate the number of moles of reactants, check if there is a limiting reagent and finally calculate the number of moles of Cu(OH)₂.

mol reactants:

Molarity = mol/L solution ⇒ mol = M x L

mol CuSO₄ = 1.50 mol/L x (150 mL/1000 mL/L) = 0.225 mol

mol KOH =  3.0 mol/L x (150 mL/1000 mL/L) = 0.45 mol

We know from the coefficients of the balanced equation that 1 mol of CuSO₄ will react with 2 mol KOH to produce 1 mol Cu(OH)₂ , thus 0.225 mol CuSO₄ will react with 0.45 mol KOH to produce 0.225 mol Cu(OH)₂.

Now we can calculate ΔHrxn:

= - 7797 J / 0.225 mol Cu(OH)₂ = -34,653J/mol Cu(OH)₂  

( Do not forget the negative sign in the change in enthalpy since we are dealing with an exothermic reaction )

yes they ca   is ur ansswerrr

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml