Question

½H2(g) + ½I2(g) → HI(g), ΔH = + 26 kJ/mole 117 kJ/mole + C(s) + 2S(s) → CS2(g) The temperature of the surroundings will:

Answer 1

½H2(g) + ½I2(g) → HI(g), ΔH = + 26 kJ/mole
117 kJ/mole + C(s) + 2S(s) → CS2(g) ??????????

Answer 2

increase or decrease?

Answer 3

in the first reaction ΔH is positive... that means heat is produced (an exothermic reaction) ... so temperature will increase.

the 2nd reaction requires energy to be put into the system.... that means heat will be absorbed (endothermic reaction) ... so, temperature will go down.

Answer 4

i would disagree with that conclusion PaxPolaris because when ever dH is positive temperature of surroundings will decrease and that kind of reactions are endothermic, where as dH is negative temperature of surroundings will increase cause heath is generated and those kind of reactions are called exothermic reactions

Answer 5

117 kJ/mole + C(s) + 2S(s) → CS2(g) <<<<< I dont understand with this one. I think u're making a mistake, can u re-write it the question more clearly?

Answer 6

Kryten is right. ΔH should be negative... in an exothermic reaction which i think the first reaction is:
H2(g) + ½I2(g) → HI(g)