Question

11. Assuming that the gases are ideal, calculate the amount of work done (joules) in each of the following reactions at 25 degrees Celsius. a. 4 HCl(g) + Ox(9) + 2Cl(g) + 2 H2O(g) b. 2 NO(g) → Na(g) + O2(g)

Answer 1

For a: Work done for the given reaction is 2477.572 J.

For b: Work done for the given reaction is 0 J

Step-by-step explanation:

To calculate the work done for the reaction, we use the equation:

`W=-P\Delta V`

Ideal gas equation follows:

`PV=nRT`

Relating both the above equations, we get:

`W=-\Delta n_gRT` ......(1)

where,

`\Delta n_g` = difference in number of moles of products and reactants =
`n_g_((products))-n_g_((reactants))`

R = Gas constant = 8.314 J/K.mol

T = temperature =
`25^oC=[273+25]K=298K`

• For a:

The chemical reaction follows:

`4HCl(g)+O_2(g)\rightarrow 2Cl_2(g)+2H_2O(g)`

`\Delta n_g=4-5=-1`

Putting values in equation 1, we get:

`W=-(-1mol)* (8.314J/K.mol)* 298K=2477.572J`

Hence, work done for the given reaction is 2477.572 J.

• For b:

The chemical reaction follows:

`2NO(g)\rightarrow N_2(g)+O_2(g)`

`\Delta n_g=2-2=0`

Putting values in equation 1, we get:

`W=-(0mol)* (8.314J/K.mol)* 298K=0J`

Hence, work done for the given reaction is 0 J.