Consider the following thermochemical equations.
Reaction ΔrH° / kJ mol−1
HBr(g) ⟶ H(g) + Br(g) 365.7
H2(g) ⟶ 2 H(g) 436.0
Br2(g) ⟶ 2 Br(g) 193.9
What is ΔrH° for the reaction below, in kJ mol-1?
HBr(g) ⟶ ½ H2(g) + ½Br2(g)
1 answer
what is 365.7 -1/2 (436.0) - 1/2 (193.9)