A 64.0 mL sample of 1.0 M NaOH is mixed with 45.0 ml of 1.0 M H_2SO_4 in a large

Question

A 64.0 mL sample of 1.0 M NaOH is mixed with 45.0 ml of 1.0 M H_2SO_4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 25.6 degree C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g degree C), and that no heat is lost to the surroundings. The Degree H rxn for the neutralization of NaOH with H_2SO_4 is -114 kJ/mol H_2SO_4. What is the maximum measured temperature in the Styrofoam cup? Consider how many moles of H_2SO_4 react and that will tell you the total amount of heat from the reaction. The heat from the reaction all ends up in the water, raising its temperature: q = mc_s Delta T

Explanation / Answer

V = 64 ml

M = 1 NaOH

V2 = 45 ml

M2 = 1 H2SO4

..........

T1 = 25.6°C

T2 =

Cp = 4.184 J/gC

2NaOH + H2SO4 --> Na2SO4(aq) + H2O H = -114 kJ/mol H2SO4

Find limiting reactant

Moles of NaOh = M*V = 1*64 = 64 mmol of NaOH

Moles of H2SO4 = M*V = 1*45 = 45 mmol of H2SO4

45 mmol of aicd will need 90 mmol of NaOH, we do not have so

H2SO4 is the limiting reactant

Therefore

n = 45 mmol of H2SO4 o

Calculate

total volume of solution

VT = V1 + V2 = 64+45 = 109 ml

MT = 1g/ml * 109 ml = 109 g

T1 = 25.6°C

Tf = ?

Q = mol H2SO4 * H rxn = n * Cp *(T2-T1)

Hrxn = -114 kJ/mol H2SO4

find total amount of mol H2SO4

mol H2sO$ = V*M = 45/1000 * 1 = 0.045 mol of acid

Hrxn = n*dH = 0.045*-114 = -5.13 kJ

Qwin = -Qlost

-Qlost = -5.13 kJ

Qwin = water heat earning

Qwin = m*Cp*(Tf-Ti)

-(-5.13 kJ( = m * Cp *(T2-T1)

5130 J = m*Cp*(T2-T1)

m = 109 g (from solution

Cp = 4.18 J/gC

T1 = 25.6

5130 J = m*Co*(T2-T1)

5130 J = 109*4.18(T2-25.6)

11.25 = T - 25.6

T = 11.25+25.6 =36.85 °C

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