Consider the following equilibria for gases : H2S(g) H2S(aq) + 4.6 Kcal ; SO2(g) SO2(aq) + 8.6 Kcal. Assuming the randomness factor to be about the same for each, which of the gases will have the lower solubility in water ?
A.H2S ; H2S loses less heat, therefore less soluble✓ Correct
B.SO2 ; H2S loses less heat, therefore more soluble
C.H2S : H2S loses more heat, therefore less soluble
D.SO2 : SO2 loses more heat, therefore less soluble
Explanation
Since, less heat is released in 1st reaction as compared to the heat released in the 2nd reaction. So, 1st reaction is less spontaneous and hence, 2 H S will be less soluble in in water.