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

Answer:

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

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

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:

$W=-(-1mol)\times (8.314J/K.mol)\times 298K=2477.572J$

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:

$W=-(0mol)\times (8.314J/K.mol)\times 298K=0J$

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