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A chemistry practice problem focused on thermochemistry, specifically calculating enthalpy change (q) and work (w) using the first law of thermodynamics. Students are given a reaction involving h2 and br2, and are asked to determine the limiting reactant, calculate the heat released, and determine the initial and final volumes to find the work. This problem is ideal for students studying thermodynamics and chemical reactions.
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= D H and
w = – P D V , where D V = the change in the volume of gas.
To determine q, determine the heat released for the quantities given. We must
determine the limiting reactant.
For H 2 :
2 molHBr
For Br 2 :
2 molHBr
1 molBr 80 .gBr 2 2
2 2 × × =
The Br 2 is the limiting reactant. The amount of heat given off when 1.0 mol HBr is
produced is determined as follows:
36 kJ 2 molHBr
Note: look at the thermochemical equation to see how the conversion factor was
obtained.
To determine w , determine the final an initial volume. This isn’t according to the
balanced equation, but according to the actual amounts of gas in the reaction chamber.
To determine this, we know we have one mole of HBr gas, but there is also some of the
unreacted H 2 gas present in the chamber as well. To determine how many moles of H (^2)
reacted, we can use the amount of HBr that is produced and use the balanced equation
as follows.
2
2
1 molH
The moles of H 2 unreacted is 3.0 mol – 0.5 mol = 2.5 mol H (^2)
Moles of gas initially are 3.00 moles (from the hydrogen gas). Moles of gas at the finish
2.5 mol (unreacted H 2 ) + 1.0 mol HBr (produced) = 3.5 moles.
Using PV = nRT , the V i and V f can be determined:
1 atm
1 atm
f =
nRT V
nRT V
i
i i
w = – P (D V ) = – P ( V f – V i ) = – 1 atm (61 L – 73 L) = – 12 L
. atm
Convert the work to units of kJ in order that the work can be added to the heat to get D E.
36 kJ ( 1.2kJ) 37 kJ
1 kJ
1 L atm
12 L atm