calculate the work done in an open vessel at 300K, when 112g iron reacts with dil. HCl .
Answers
Answered by
30
The reaction involved is:
Fe+2HCl→FeCl2+H2
Work done=p(V2-V1)
Since mole of Fe used=112/56=2
The moles of H2 formed =2
P*(VH2-Vi)=P*VH2
As Vi=0
For H2 P*VH2=nRT
VH2=nRT/P
W=P*nRT/P=nRT=2*2*300=1200 cal or 1.2kcal
Fe+2HCl→FeCl2+H2
Work done=p(V2-V1)
Since mole of Fe used=112/56=2
The moles of H2 formed =2
P*(VH2-Vi)=P*VH2
As Vi=0
For H2 P*VH2=nRT
VH2=nRT/P
W=P*nRT/P=nRT=2*2*300=1200 cal or 1.2kcal
kvnmurty:
we usualyl express heat /energy in calories... work done in Joules perhaps ? as it is SI unit.
Answered by
0
Answer:
1.2kcal
Explanation:
The reaction involved is:
Fe +2HCI ------>FeCl12 +H2
Work done = P(V2-V1)
Since mole of Fe used =112/56 =2
The moles of H2 formed =2
P*(VH2-Vi) = p*VH2
As Vi=0
For H2 P* VH2 = nRT
VH2 = nRT/P
W =P*nRT/P 1.2 kcal = nRT = 2*2*300=1200 cal or 1.2kcal
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