At constant pressure, which of these systems do work on

hionormf

hionormf

Answered question

2021-12-15

At constant pressure, which of these systems do work on the surroundings? Check all that apply.
a) 2A(g)+3B(g)4C(g)
b) 2A(g)+3B(g)4C(g)
c) A(g)+B(g)3C(g)
d) A(s)+2B(g)C(g)

Answer & Explanation

chumants6g

chumants6g

Beginner2021-12-16Added 33 answers

Step 1
Consider the following reaction a:
The number of moles of gaseous reactants (nr) is:
nr=3+2
=5
The number of moles of gaseous products (np) is 4.
Now, calculate Δng for this reaction as:
Δng=npnr
Substitute 4 for np and 5 for nr
Δng=45
=1
The work done is calculated as:
w=ΔngRT
Substitute -1 for Δng
w=(1)RT
=+RT
Therefore, work is not done by the system (reaction a) on the surroundings.
Consider the following reaction d:
The number of moles of gaseous reactants (np) is 2.
The number of moles of gaseous products (np) is 1.
Now, calculate Δng for this reaction as:
Δng=npnr
Substitute 1 for np and 2 for nr
Δng=12
=1
The work done is calculated as:
w=ΔngRT
Substitute -1 for Δng
w=(1)RT
=+RT
Therefore, work is not done by the system (reaction d) on the surroundings.
Step 2
Consider the following reaction b:
The number of moles of gaseous reactants (nr0 is 1
The number of moles of gaseous products (np) is 2.
Now, calculate Δng for this reaction as:
Δng=npnr
Substitute 2 for np and 1 for nr
Δng=21
=+1
The work done is calculated as:
w=ΔngRT
Substitute +1 for Δng
w=(+1)RT
=RT
Therefore, work is done by the system on the surroundings.
Step 3
Consider the following reaction c:
The number of moles of gaseous reactants (np) is:
nr=1+1
=2
The number of moles of gaseous products (np) is 3.
Now, calculate Δng for this reaction as:
Δng=npnr
Substitute 3 for np and 2 for nr
Δng=32
=+1
The work done is calculated as:
w=ΔngRT
Substitute +1 for Δng

Kindlein6h

Kindlein6h

Beginner2021-12-17Added 27 answers

Step 1
When a gas is evolved during a chemical reaction, the gas can be imagined as displacing the atmosphere - pushing it back against the atmospheric pressure. The work done is therefore V×P where V is the volume of gas evolved, and P is the atmospheric pressure.
A(g)+B(g)C(g)
In this reaction 2 volumes of reagent combine to form 1 vol of product. The total volume falls, so the work done is positive (the atmosphere does work on the gaseous system).
2A(g)+2B(g)5C(g)
4 vols of reagent combine to form 5 vols of product. The total volume increases and negative work is done.
A(s)+B(s)C(g)
we can assume that the volume of gas produced exceeds the volume of solid consumed, so again negative work is done.
2A(g)+B(g)C(g)
In this reaction 3 volumes of reagent combine to form 1 vol of product. The total volume falls, so the work done on the atmosphere is positive.

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