What is the enthalpy change when 12.8g H2(g) reacts with excess Cl2(g) to form HCl(g).
the equation
ΔH∘rxn=∑(n×Δf products)−∑(m×ΔHreactants)
, wheren, m
- the number of moles of each product and reactant, respectively
= 12.8/2.016
= 6.35 moles
∆H of Hydrogen gas = 0
∆H for Cl = 127
= 6.35(0) - 127(1)
= 0 -127
= 127J/mole
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