Question

Iron reacts with dil.HCI according to the equa-
tion Fe(s) + 2HCl(aq)= FeCl2(aq) +H2 (g).
The work done when 2.8 g of iron reacts with excess
of dil.HCI in a closed vessel at 27° C is
→ FeCl2 (aq)
+ H2(g)-- The
1) -124.7 J
2) +124.7 J
3) Zero
4) - 138.2 J​

Answer 1

Explanation:

Solution:- (D) 0,−600

In a closed vessel →

When the volume is fixed,

W=−PΔV

∵ΔV=0

⇒W=0

Hence for the reaction in closed vessel of fixed volume, work done will be zero.

In an open vesel →

Molar mass of Fe=56g

Given mass of Fe=56g

As we know that,

No. of moles=Molar massMass

Therefore, number of moles of Fe used-

n=5656=1 moles

As we know that,

w=−PΔV=−nRT[∵PV=nRT]

Given T=300K

∴w=−1×2×300=−600 cal

Hence for the reaction in an open vessel, the work done will be −600 cal.

Answer 2

Explanation:

Calculate the work done when 56g of iron reacts with hydrochloric acid