Question

For the reaction H${}_2$ (g) + I${}_2$(g)$\rightleftharpoons$2HI(g),K${}^o$${}_c$= 66.9 at 350${}^{\circ }$C and K${}^o$${}_c$= 50.0 at 448 ${}^{\circ }$C. The reaction has

• A. $\Delta$H = +ve.
• B. $\Delta$H = -ve.
• C. $\Delta$H = Zero
• D. $\Delta$H whose sign cannot be predicted

Answer 1

The correct option is B

$\Delta$H = -ve.

Since $K_c$ decreases with increase in temperature, the reaction must be exothermic. Hence $\Delta$H = -ve.