Question

Work done when 65.38g of zn is dissolved completely in hcl in an open beaker at 300k is

Answer 1

Answer:- work = -2496 J

Solution:- The balanced equation for the reaction of zinc with HCl is:

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

Given grams of Zn are converted to moles. Moles of hydrogen gas formed will also be equal as there is 1:1 mol ratio between zinc and hydrogen.

$65.38gZn(\frac{1molZn}{65.38gZn})(\frac{1molH_2}{1molZn})$

= $1molH_2$

From first law of thermodynamics, $w=-p\Delta V$  .....(1)

From an ideal gas law equation, PV = nRT    

So, $P\Delta V=\Delta nRT$    ---------(2)

From equation 1 and 2:

$w=-\Delta nRT$

From above calculation, $\Delta n$ is 1 mol.

T = 300 K

R = $0.0821\frac{atm.L}{mol.K}$

Let's plug in the values in the equation:

$w=-1mol*0.0821\frac{atm.L}{mol.K}*300K$

w = -24.63 atm.L

Since, 1 atm.L = 101.325 J

So,  $w=-24.63atm.L(\frac{101.325J}{atm.L})$

w = -2495.63 J or it is almost -2496 J

The negative sign indicates the work done by the system which is also clear as the gas is formed means there is expansion in volume.

So, the work done is -2496 J.