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3 years ago

I am having the worst day ever, Someone please cheer me up. :( I'll mark you as brainiest. :(

Chemistry

1 answer:

Olegator [25]3 years ago

5 0

Once I told a chemistry joke

I got no reaction.

What do you do with a sick chemistry teacher?

If you can't helium, and can't curium, you might as well barium.

I don't trust atoms,

I heard they make up everything

Hope I helped make ur day. :)

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3 years ago

11. Assuming that the gases are ideal, calculate the amount of work done (joules) in each of the following reactions at 25 degre

gayaneshka [121]

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.

3 0

3 years ago