a penalty 10 Question (5 points) O See page 236 7th attempt Al See Periodic Tabl

Question

a penalty 10 Question (5 points) O See page 236 7th attempt Al See Periodic Table O See Hint A 64.0 ml sample of 1.OM NaOH is mixed with 47.0 ml of 1.0 MHSO, 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 226°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 1.0 g/ml, the specific the mixed J/g °C, and no is to the surroundings. The Allen for the that heat of solutions is 18 that heat lost 4 neutralization of NaOH with HSO, is -114 KJ/mol HSo, What is the maximum measured temperature in the Styrofoam cup? 3046 6th attempt VIEW SOLUTION 8 OF 10QUESTIONS COMPLETED

Explanation / Answer

Moles of NaOH= molarity* volume in L=1*64/1000=0.064

Moles of H2SO4= 1*47/1000 =0.047

The reaction between NaOH and H2SO4 is

2NaOH + H2SO4-------àNa2SO4+ 2H2O

Molar ratio of NaOH: H2SO4=2:1

Actual ratio = 0.064:0.046= 1.5:1

So limiting reactant is NaOH.

Moles of H2SO4 consumed = 0.064/2=0.032

Heat of reaction =-114 Kj/mole

Hence heat liberated from 0.032 moles of H2SO4

Enthalpy change= -114*0.032 KJ=-3.648 KJ=3648j

This much heat released is taken by the solution.

Volume of solution =64+47=111ml,

Mass of solution = volume*density =111*1= 111gm

Specific heat= 4.184 J/gm.deg.c

Hence heat released= mass of solution* specific heat* temperature change

3648= 111*4.184*temperature rise

Temperature rise= 3648/(111*4.184)=7.85

T2-22.6= 7.85, T2= temperature after the reaction.

Temperature =7.85+22.6= 30.45 deg.c

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