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Tanzania [10]
3 years ago
5

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with

12.0 moles of H2 to form NH3, calculate the work done (in joules) against a pressure of 1.0 atm at 25°C.
Chemistry
1 answer:
Sphinxa [80]3 years ago
6 0

Answer : The work done is, 1.98\times 10^4J

Explanation :

The given balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

When 4 moles of N_2 react with 12 moles of H_2 then it gives 8 moles of NH_3

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of N_2 + Moles of H_2

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of NH_3

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 25^oC=273+25=298K

Putting values in above equation, we get:

1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K

V=195.7L

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V

where,

w = work done

p = pressure of the gas = 1.0 atm

\Delta V = change in volume = -195.7 L

Now put all the given values in the above formula, we get:

w=-p\Delta V

w=-(1.0atm)\times (-195.7L)

w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, 1.98\times 10^4J

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