Thermodynamics chemical reaction problem

AI Thread Summary
The discussion revolves around calculating the work (w) and heat (q) for the reaction of sodium and chlorine gas to form sodium chloride at standard conditions. The user outlines the reaction and attempts to calculate the change in moles of gas, finding a delta(n(g)) of -1.16, leading to a calculated work of 2.87 kJ. However, this value does not match the textbook answer, prompting a consideration of whether the volume change can be applied using the formula w = pΔV. The conversation highlights the complexities of thermodynamic calculations in chemical reactions, particularly regarding gas behavior and heat transfer.
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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.
 
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