Question

What is the enthalpy change when 12.8g H₂(g) reacts with excess Cl₂(g) to form HCl(g).

Answer 1

 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