Thermodynamics chemical reaction problem

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SUMMARY

The discussion focuses on calculating the work (w) and heat (q) for the reaction of sodium (Na) and chlorine gas (Cl2) to produce sodium chloride (NaCl) at standard conditions (1 atm, 298 K). The user applied the formula w = -delta(n(g)) * R * T, determining delta(n(g)) to be -1.16, resulting in a calculated work of 2.87 kJ. However, this result was questioned due to the volume change from gaseous Cl2 to solid NaCl, prompting a consideration of using w = pΔV for a more accurate assessment.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically work and heat in chemical reactions.
  • Familiarity with the ideal gas law and molar volume of gases.
  • Knowledge of the first law of thermodynamics.
  • Ability to perform stoichiometric calculations involving moles and mass.
NEXT STEPS
  • Research the application of the equation w = pΔV in thermodynamic calculations.
  • Study the relationship between enthalpy (q) and work (w) in chemical reactions.
  • Explore the concept of standard state conditions and their impact on reaction calculations.
  • Learn about the heat capacities of reactants and products to calculate q accurately.
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Chemistry students, educators, and professionals involved in thermodynamics and chemical reaction engineering will benefit from this discussion.

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Homework Statement


What are q and w (in kJ) for the reaction of 22 g of Na(s) and 82 g of Cl2(g) to produce NaCl at 1 atm and 298 K.


Homework Equations



w=-delta(n(g))* R*T where delta(n(g)) = the moles of gas created during the reaction

The Attempt at a Solution



First, I wrote the chemical reaction for Na and Cl2

2Na(s) +Cl2(g)=> 2NaCl

delta(n(g))=0-1 = -1

82 g Cl2*(1 mol Cl2/70.98 g Cl2)*(-1 mol gas/1mol Cl2)= -1.16

delta n(g) =-1.16
w=-delta(n(g))*RT=(-1.16)(8.314)(298 K)=2.87 kJ

But that's not the answer according to the back of my textbook.
 
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The volume is changing from the volume of Cl2 gas to essentially zero.

Can one apply w = p \Delta{V}? 1 mole of gas has a volume of 22.4 l.

Then there is the heat.
 
Last edited:

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