Reaction 1: MgO(s) + 2HCl(aq) ? MgCl2(aq) + H2O(l); ?H1 Reaction 2: Mg(s) + 2HCl
ID: 759102 • Letter: R
Question
Reaction 1: MgO(s) + 2HCl(aq) ? MgCl2(aq) + H2O(l); ?H1 Reaction 2: Mg(s) + 2HCl(aq) ? MgCl2(aq) + H2(g); ?H2 Reaction 3: H2(g) + 1/2O2(g) ? H2O(l); ?H3 = -285.8 kJ/mol a) What is the enthalpy per mole of MgO consumed if the reaction of 0.545g of MgO with HCl caused the temperature of the water in the calorimeter to go from 23.0?C to 25.8?C? (Assume Ccal = 418J/?C) b) What is the enthalpy per mole of Mg(s) consumed if the reaction of 0.356g of Mg metal with HCl caused the temperature of the water in the calorimeter to go from 23.0?C to 36.5?C? (Assume Ccal = 418J/?C)Explanation / Answer
To calculate the enthalpies of reaction, in order to solve questions (2) and (4) you need to know the heat evolved from reaction. You can determine this by using a calorimeter with water where the heat evolved from reaction is transferred to water. You also need to know the mass of water used in the calorimeter and the change in temperature. You seem to have the T values but you don't provide the mass of water used. The specific heat of water is 4.184 J/mol*K, thus the heat would be calculated as: heat = (mass H2O) x (specific heat) x (change in T) Once you determine the number of Joules released, all you need to do is to divide by the number of moles of Mg or MgO used in the reaction to get the heat of reaction. In order to solve question (1) and (3) you have to determine the number of moles of reagent used. Since you didn't specify the concentration of HCl I'll assume you used excess for this reaction; therefore, the limiting reagents will be Mg and MgO respectively and the heat transferred to the solvent will be: (1) heat transferred to water = (0.3528 g Mg) x [(1 mole)/(24.31 g)] x (-412.19 kJ/mol) = -5.982 kJ (3) heat transferred to water = (1.11 g MgO) x [(1 mole)/(40.30 g)] x (-130.65 kJ/mol) = -3..60 kJ To determine the enthalpy of reaction of Mg + 1/2 O2(g) ---> MgO(s) you need two add-up the following processes: Mg(s) + 2 HCl(aq) ----> MgCl2(aq) + H2(g).............H = -412.19 kJ/mol MgCl2(aq) + H2O(l) ----> MgO(s) + 2 HCl(aq).........H = 130.65 kJ/mol H2(g) + 1/2 O2(g) ----> H2O(l)...............................H = -285.8 kJ/mol --------------------------------------… Mg(s) + 1/2 O2(g) -----> MgO(s)............................H = -567.3 kJ/mol Which is the value you gave us at the very beginning. Notice that the third equation is nothing more than the enthalpy of formation of H2O(l).
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