Thermochemistry Practice: Calculating Enthalpy Change & Work with First Law, Schemes and Mind Maps of Chemistry

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.

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 08/05/2022

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Chemistry Practice Problems
Thermochemistry: Enthalpy change and the First Law of
Thermodynamics
Solutions:*
1. The first law of thermodynamics provides that DE= q + w. At constant pressure, q
= DH and
w = PDV, where DV = 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 H2:
HBr mol 0.6
H mol 1
HBr mol 2
H mol 0.3
2
2
=×
For Br2:
HBr mol 0.1
Br mol 1
HBr mol 2
Br g 8.159
Br mol 1
Br g .80
22
2
2
=××
The Br2 is the limiting reactant. The amount of heat given off when 1.0 mol HBr is
produced is determined as follows:
kJ 36
HBr mol 2
kJ 4.72
Br mol 0.1
2
=
×
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 H2 gas present in the chamber as well. To determine how many moles of H2
reacted, we can use the amount of HBr that is produced and use the balanced equation
as follows.
2
2
H mol 50.0
HBr mol 2
H mol 1
HBr mol 0.1 =×
pf2

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Chemistry Practice Problems

Thermochemistry: Enthalpy change and the First Law of

Thermodynamics

Solutions:

  1. The first law of thermodynamics provides that D E = q + w. At constant pressure, q

= 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 :

  1. 0 mol HBr 1 molH

2 molHBr

  1. 0 molH 2

2 × =

For Br 2 :

  1. 0 mol HBr 1 molBr

2 molHBr

  1. 8 gBr

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

  1. 4 kJ
  2. 0 molBr 2 =−

×

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. 50 molH 2 molHBr

1 molH

  1. 0 molHBr× =

Chemistry Practice Problems

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:

61 L

1 atm

  1. 5 mol(0.0820 6 L atm/mol K)(298K)

73 L

1 atm

  1. 0 mol(0.0820 6 L atm/mol K)(298K)

f =

P

nRT V

P

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. 2 kJ 1000 J

1 kJ

1 L atm

101.3J

12 L atm

× =−

− ⋅ ×

E