In the reaction: H2+I2=2HI In a 2 L flask 0.4 moles of each H2 and I2 are taken. At equilibrium 0.5 moles of HI are formed. What will be the value of equilibrium constant Kc ?

Question

In the reaction: H2+I2=2HI In a 2 L flask 0.4 moles of each H2 and I2 are taken. At equilibrium 0.5 moles of HI are formed. What will be the value of equilibrium constant Kc ? (A) 20.2 (B) 25.4 (C) 0.284 (D) 11.1

Tardigrade Admin · Accepted Answer

underset textAt textequilibrium =0.15 mathop underset 0.4-0.25 mathop underset 0.4 mathop H2 + underset= 0.15 mathop underset0.4-0.25 mathop underset0.4 mathopI2 = underset0.05 mathop underset0 mathop2HI Kc=([HI]2/[H2][I2]) =(( (0.50/2) )2/( (0.50)2 )( (0.50)2 )=(0.5× 0.5/0.15× 0.15)=11.11

Q. In the reaction : $H_{2} + I_{2} = 2 H I$ In a 2 L flask 0.4 moles of each $H_{2}$ and $I_{2}$ are taken. At equilibrium 0.5 moles of HI are formed. What will be the value of equilibrium constant $K_{c}$ ?

20.2

25.4

0.284

11.1

$At equilibrium = 0.15 0.4 - 0.25 0.4 H_{2} + = 0.15 0.4 - 0.25 0.4 I_{2} = 0.05 0 2 H I$ $K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$ $= \frac{( \frac{0.50}{2} ) ^{2}}{( \frac{0.50}{2} ) ( \frac{0.50}{2} )} = \frac{0.5 \times 0.5}{0.15 \times 0.15} = 11.11$

Q. In the reaction : H2​+I2​=2HI In a 2 L flask 0.4 moles of each H2​ and I2​ are taken. At equilibrium 0.5 moles of HI are formed. What will be the value of equilibrium constant Kc​ ?