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