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Homework Help: Chemistry - Limiting Reactants, Thermochemical Change

  1. Apr 25, 2014 #1
    The problem is that 'my' solution is at variance with the solution provided for the question in the solutions manual that accompanies the textbook I am using.

    1. The problem statement, all variables and given/known data
    "A 14.1-mL sample of 0.996 M NaOH is mixed with
    32.3 mL of 0.905 M HCl in a coffee-cup calorimeter (see Section
    6.6 of your text for a description of a coffee-cup calorimeter).
    The enthalpy of the reaction, written with the lowest
    whole number coefficients, is 55.8 kJ. Both solutions are at
    21.6°C prior to mixing and reacting. What is the final temperature
    of the reaction mixture? When solving this problem, assume
    that no heat is lost from the calorimeter to the surroundings,
    the density of all solutions is 1.00 g/mL, the specific
    heat of all solutions is the same as water, and volumes are

    2. Relevant equations
    Besides the conversion factors,
    q(absorbed) = -q(released)
    q = specific heat capacity * mass * Δtemperature

    3. The attempt at a solution

    The final temperature is the temperature recorded when thermochemical equilibrium is attained.

    Given the 1.00 mL:1.00 g conversion, it means that the entire mass of the solutions is 46.4 g.

    The thermochemical system is the NaCl and the H2O, which releases 55.8 kJ of heat when formed from 1 mol of NaOH and 1 mol of HCl.

    The reaction is as follows: NaOH + HCl → NaCl + H2O: ΔH = -55.8 kJ.

    The reaction is ushered in as the result of a mixing of 14.1mL of a 0.996M NaOH solution, or

    14.1mL NaOH sol.[itex]\frac{0.996 mols NaOH}{1000mL NaOH sol.}[/itex] = 0.0140 mols NaOH,

    and a 32.3mL HCl solution of 0.905M, or

    32.3mL NaOH sol.[itex]\frac{0.905 mols HCL}{1000mL HCl sol.}[/itex] = 0.0292 mols HCl,

    There is a limiting reactant here, and so given that 0.0140 mols NaOH reacts, the amount of heat released will be equivalent to (0.0140/1)(55.8 kJ) = 0.781 kJ or 781 J.

    Easy stuff so far.

    The formation of 0.0140 mols NaCl and 0.0140 mols H2O releases 781 J to the surroundings. By the equation listed above,

    q(released)(-1) = q(absorbed)
    781 J = (massunreacted solution)(4.18J/g*°C)(Tfinal - 21.6°C)
    Tfinal = [781 J]/[(4.184J/g*°C)*massunreacted solution] + 21.6°C

    If all the NaOH reacts with 0.0140 mols of HCl to form 0.0140 mol of NaCl and 0.0140 mol of H2O to form 1.08g of H2O and NaCl, which is part of the same thermochemical system, the only place the heat can go is to the unreacted solution of the HCl, this much of which remains:

    Massunreacted solution = Mass of solutions before reaction - (Mass of H2O + Mass of NaCl) = (32.3 + 14.1)g - 1.08g = 45.3g.

    Substituting, we get

    Tfinal = [itex]\frac{781 J}{45.3g*(4.18J/g*°C)}[/itex] + 21.6°C
    = 25.7°C

    as the 'final temperature'.

    The solutions manual, in deriving the answer, for Tfinal, divides 781 J by the specific heat capacity of the solution , the same as that of water (see the question), and the total mass of the solutions before reaction, 46.4 g. This doesn't seem correct to me. Accordingly, their solution is 25.6°C (not much of a difference). I'm mainly concerned about whether my method is correct, because to not discern between system and surroundings in such a context is to miss a big one.

    So: Is my method correct?
  2. jcsd
  3. Apr 25, 2014 #2


    User Avatar

    Staff: Mentor

    Solutions manual is right. You can't heat up just part of the mixture, you heat up everything - so you heat up whole 46.4 g.

    What is - in your version - final temperature of the 1.08 g of water produced?
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