Calculate the work done when $65.38 g$ of zinc dissolves, hydrochloric acid in an open beaker at $300 K$ . (At mass of $Z n = 65.38$ )

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Solution

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 the 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:
$w=-1mol0.0821\frac{atm.L}{mol.K}300K$
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.

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