Question

The work done in an open vessel at $300$K, when $112g$ iron reacts with dilute $HCl$ to give $FeC{l}_2$, is nearly:

• A. $1.1$ kcal
• B. $0.6$ kcal
• C. $0.3$ kcal
• D. $0.2$ kcal

Answer 1

Answer: (A)

$1.1$ kcal

The reaction involved is:

$Fe+2HCl\to FeC{l}_2+H_2$

Atomic mass of $Fe=56g/mol$

Thus, 56 g of Iron reacts with 2 moles of HCl to give one mole of Hydrogen gas.

Initial volume of $H_2$ gas = $V_1=0$

Final volume of $H_2$ gas=$V_2$

Using ideal gas law:$PV=nRT$

where n=mass/molar mass, R=8314 J/K/mol and given that T=300 K

$P{V}_2=\left(112/56\right)\times 8.314\times 300=4988.4J$

Work done $=-P\Delta V=-P(V_2-V_1)=-P{V}_2=-4988.4J$

negative work done is work of expansion.

since 4184 J=1 kcal

thus $4988.4J=1.19kcal$ of work is done by the system.