When Fe(s) is dissolved in an aqueous HCl in a closed rigid vessel, the work done is

zero.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Positive

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Negative

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C zero.
Reaction:

$F e (s) + 2 H C l (q a) \to F e C l_{2} (a q) + H_{2} (g)$

but as vessel is rigid so volume change will be zero and work done will be zero.

Suggest Corrections

Similar questions

F e (s)

H C l

F e (s)

11.2 g

H C l

25^{o} C

1

2 mol of zinc is dissolved in $H C l$ at $25^{\circ} C$ . The work done in an open vessel is:

2 m o l e

H C l

25^{o} C

Join BYJU'S Learning Program