Question

Consider the following balanced thermochemical equation, sometimes used for H2S production.

1/8 S8(s) + H2(g) H2S(g) ΔHrxn = -20.2 kJ

(a) Is this an exothermic or endothermic reaction?

(b) What is ΔHrxn for the reverse reaction?
kJ

(c) What is ΔHrxn when 1.5 mol S8 reacts?
kJ

(d) What is ΔHrxn when 22.2 g S8 reacts?

kJ

Answer 1

(a) H of reaction is -20.2 KJ negative sign indicate that heat is evolved in the reaction therefore reaction is exothermic.

(b) wright down reaction in revesrse manner then equation become

H₂S_(g) 1/8 S_(s) + H_2(g) ΔHrxn = +20.2 kJ

ΔHrxn = +20.2 kJ for reverse reaction

(c) for 1/8 mole of S₈ ΔHrxn = -20.2 kJ then for 1.5 mole of ΔHrxn = (-20.8 1.5) /(1/8) = -249.6 KJ

ΔHrxn for 1.5 mole of S₈ = -249.6 KJ

(d) molar mass of S₈ = 256.52 gm /mole then 22.2 gm of S₈ = 22.2 / 256.52 = 0.0865 mole

for 1/8 mole of S₈ ΔHrxn = -20.2 kJ then for 0.0865 mole of  ΔHrxn = (-20.8 0.0865) /(1/8) = -14.3936KJ

ΔHrxn for 22.2 of S₈ = -14.3936 KJ