Question

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.

Answer 1

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, $1.98\times 10^4J$