Question

the value of the equilibrium constant for the reaction H2(g)+I2(g)2HI(g) at 720K is 48. What is the value of the equilibrium constant for the reaction 1/2H2(g)+1/2I2(g)HI(g)

Answer 1

√48 = 4√3

H₂ (g) + I₂ (g)  ⇔  2 HI (g)   at 720 °K

{ HI }  means the concentration, activity, mass fraction, amount fraction, molality etc. of  HI.

$K_c=\frac{\{HI\}^2}{\{H_2\}\ \{ I_2\}}=48\\\\\frac{1}{2}H_2(g)+\frac{1}{2}I_2 (g) \ \textless \ =\ \textgreater \ HI(g)\\\\K_c_1=\frac{\{HI\}}{\{H_2\}^\frac{1}{2}\ \{ I_2\}^\frac{1}{2}}=\sqrt{K_c}\\\\=\sqrt{48}=4\sqrt3$

Answer 2

Dont forgot to like.