For the reaction: H2(g) +I2(g) leftharpoons 2HI(g), the standard free energy is Δ G° 0. The equilibrium constant (K) would be.

Question

For the reaction: H2(g) +I2(g) leftharpoons 2HI(g), the standard free energy is Δ G° > 0. The equilibrium constant (K) would be. (A) K=0 (B) K > 1 (C) K=1 (D) K < 1

Tardigrade Admin · Accepted Answer

Δ G°=-RT lnK If lnK=-ve then only Δ G°=+ve (or > 0). Thus, K < 1

Q. For the reaction : $H 2_{(g)} + I 2_{(g)} ⇌ 2 H I_{(g)}$ , the standard free energy is $Δ G^{\circ} > 0$ .
The equilibrium constant $(K)$ would be________.

$K = 0$

$K > 1$

$K = 1$

$K < 1$

$Δ G^{\circ} = - RT l n K$
If $l n K = - v e$ then only $Δ G^{\circ} = + v e$ (or $> 0$ ).
Thus, $K < 1$

Q. For the reaction : H2(g)​+I2(g)​⇌2HI(g)​, the standard free energy is ΔG∘>0. The equilibrium constant (K) would be________.

K=0

K>1

K=1

K<1