Question

Input Data Reactant Mass (9) .5000 1.00 Mg Volume (mL) 100.0 100.0 AT (C") 19.7 Mgo 9.0 Calculated Enthalpies of Reaction Mg(s) + 2 HCl(aq) --> MgCl2(aq) + H2(g) AH = -95.76 kcal/mol MgO(s) + 2 HCl(aq) --> MgCl2(aq) + H2O(1) AH = -36.27 kcal/mol Problem: Solve for the following enthalpy changes 1) When 6.73 g NaOH(s) is reacted with 200.0 mL of HCl(aq), the temperature increases 20.10 Celsius degrees NaOH(s) + H"(aq) --> Na"(aq) H2O(0) AH = ? 2) When 3.62 g NaOH(s) is dissolved in 200.0 mL of water, the temperature increases 4.68 Celsius degrees NaOH(s) --> Na"(aq) + OH"(aq) AH = ?

Answer 1

Ans

1. - 23.89 kcal/ mole

2. - 1.034 kcal/ mole

Explained below

NaOH (5) + H+ (aq) → Nat (aq) + H2011. 6.739 Naolt = $6.739 = 0.16825 mall 40g/mol specific heat of water a I calorie / gram Mass of watch solution - 200g considering on density Iglml] so, Heat absorbed by watch to increase temperature by 20.10€ = 200 g x 1 caloriel g * 20.10°c - 4020 calorie : 4.020 kcal en kool heat in aenerated by reaction of 0.16825 mole of NaoH so, OH = -4.02 okcal |-23.89 kcal/mole 0.016825 mele [negative sign because heat is generated - so, the reaction is enothermic I NaOH(s) Nat(aq) + OH(aq) Mole of Naot = 3.62 g 0.0905 mole og/mole Heat generated a 200 g x leal/g x 4.68° - 936 cal. so.936 kcal 0.905 mole € 1.034 kcal/more. Negative sign for enothermic reaction. sc, OH = -0.936 kcal