H2(g) + Cl2(g)  2HCl(g) , ΔH = -184kJ ,if 2 moles of H2 reacts with 2 moles of Cl2 ,then ΔU is equal to-

Solution -
H2(g) +Cl2(g) → 2HCl(g)      [ ∆H = -184KJ]
 For reaction of  2 moles of reactants then,

2H2(g) +2Cl2(g)  → 4HCl(g)  [∆H = 2×(-184) = -368 KJ]
 Also,
: ∆H =∆U +∆ngRT
∆ng = no. Of moles of gaseous product - no. Of gaseous reactants
=4-4 =0

Putting in formula,
∆H = ∆U + ∆ngRT
-368 = ∆U + (0 ×RT)
∆U = -368KJ

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