Question

Which of the following is/are correct ?

• A. When 1 mole of Zn is dissolved in excess HCl the work done is approximately equal to –2.46 kJ in open beaker at 300 K and 1 atm.
• B. When 1 mole of Zn is dissolved in excess HCl work done is equal to zero in closed beaker
• C. Both (1) and (2) are correct
• D. Neither (1) and nor (2) are correct

Answer 1

Answer: (C)

Both (1) and (2) are correct

Answer 2

Zn + 2HCl ¾® ZnCl₂ + H₂

The no. of moles of H₂ produced = 1mole

Vol. of H₂(g) produced = $\frac{0.082 \times 300}{1}$L = 24.6 L = 24.6 × 10^–3m³

\ in open beaker, DW = – P_ext × DV

= – 10⁵ × 24.6 × 10^–3 = – 2460 J

In a closed vessel , DV = 0 and DW = 0