4. What will be the effect on the enthalpy change determined if the true concent
ID: 698406 • Letter: 4
Question
4. What will be the effect on the enthalpy change determined if the true concentration of the NaOH was 5% lower than that stated on the label. (2 pts) a. b. HCl were 596 Higher than that stated on the label. (2 pts) 5. A 50.0mL sample of a 1.00 M solution of MgSO, is mixed with 50.0mL of 2.00 M NaOH in a calorimeter. The temperature of both solutions was 23.2°C before mixing and 29.3°C after mixing. The heat capacity of the calorimeter is 15 JPC. From these data, calculate H for the process Mgso,aq) + 2NaOH(aq) Mg(OH),(') + Na,SO4(ag) Assume that the specific heat and density of the solution after mixing are the same as those of pure water and that the volumes are additive. (4 pts) honor I have neither given nor received unauthorized support on thisExplanation / Answer
when the concentration of NaOH was 5% lower than the table value, the actual moles of NaOH will be less than the cakculated value. since enthalpy is expressed in terms of KJ/mole, the actual moles are less and this leads to higher enthalpy due to a reduction in actual moles of NaOH.
when the concentration of HCl is more than the label valuel, the opposite effect can be seen and enthalpy of vaporization calculated is less.
2. Volume of solution = 50+50= 100ml, density of solution = volume*density of water= 100*1g/ml=100gm
specfiic heat of solution = 4.184 J/gm.deg.c ( same as water)
enthalpy change of solution =mass of solution* specific heat* temperature change = 100*4.184*(23.2-29.3) =-2552 joules
heat added to calorimter = heat capacity of calorimete* temperature change = 15*(23.2-29.3) =-91.5 joules
overall heat added = enthalpy change due to reaction = -2552-91.5=-2643.5 joules
moles of MgSO4= molarity* volume in L= 1*50/1000
enthalpy change/mole of MgSO4= -2643.5*1000/50 J/mole =-52870 J/mole= -52.87KJ/mole
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