Question

(a) If 913 J of heat are transferred into the system, and the system does 157 J of work, compute the change in the internal energy, and interpret the sign of your result.

                             (b) Meals-Ready-to-eat (MREs) are military meals that can be heated on a flameless heater via the reaction Mg(s) + 2 H₂O(l) --- > Mg(OH)₂(s) + 2 H₂(g)

                 (i) Calculate ∆Hº_298,rxn

                        (ii) Calculate the number of grams of Mg(s) needed for this reaction to release enough energy to increase the temperature of 75mL of water from 21 to 79 ^oC. Assume d_water = 1 g/mL. Note: heat given off is proportional to the moles of Mg(s).

Answer 1

(a) From first Law of thermodynamics ,

Q + W = dE

(where Q = Heat transferred to System, W = work done by system ; dE = change in internal energy ).

Q = 913 J ; W = -157 J (as per IUPAC convention work done by system is expressed as -ve sign )

So, dE ( change in internal energy) = 913 J - 157 J = +756 J

+ve sign of change in internal energy indicates that Internal energy of System have increased, part of heat transferred to system is absorbed as internal energy.

(b)We have following reaction ;

Mg(s) + 2 H₂O(l) $\rightarrow$ Mg(OH)₂(s) + 2 H₂(g)

i.   ∆_rHº₂₉₈ of reaction :

∆_rHº₂₉₈ = _Products ΣHº_m − _Reactants ΣHº_m

∆_rHº₂₉₈= {Hº_m( Mg(OH)₂(s)) + 2*Hº_m(H₂(g))} − {Hº_m( Mg(s)) + 2*Hº_m(H₂O(l) )}

Hº_m is the standard molar enthalpy of a species.

The standard enthalpies of formation of elements in their reference states are zero, thus

Hº_m( Mg(s) = Hº_m(H₂(g) = 0

we have (various sources) : Hº_m(H₂O(l) = −285.83 KJ /mol

Hº_m( Mg(OH)₂(s) = −924.7 KJ / mol

putting values,

  ∆_rHº₂₉₈= { −924.7 KJ / mol+ 0} − {0+ 2* −285.83 KJ /mol} = -353.04 KJ/mol

(-ve sign indicates exothermic reaction )

ii) Grams of Mg(s) needed for this reaction to release enough energy to increase the temperature of 75mL of water from 21 to 79 ^oC

Specific heat capacity of water(C) = 1 cal /g-deg. = 4.184 J / g-deg

Mass of water = 75 g

Amount of heat required, Q = m*dT * C = 75 g *58 deg* 4.184 J / g-deg = 18.2 KJ

Since reaction of 1 mol of Mg(s) releases = 353.04 KJ

Thus moles of Mg(s) needed = 18.2 KJ / 353.04 KJ =0.052 moles

( Molar mass of Mg = 24.3 g /mol )

or   Grams of Mg(s) needed =  0.052 moles * 24.3 g /mol = 1.26 g