15 moles of H2 and 5.2 moles of I2 are mixed and allowed to attain equilibrium at 500°C. At equilibrium, the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

Question

15 moles of H2 and 5.2 moles of I2 are mixed and allowed to attain equilibrium at 500°C. At equilibrium, the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is (A) 50 (B) 15 (C) 100 (D) 25

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/d1186c3a4039fe0652a1dbc8083e8117-.png" /> Equilibrium constant (Kc) =(( HI )2/[ H 2][ I 2]) =(10 × 10/10 × 0.2) =50

Q. $15$ moles of $H_{2}$ and $5.2$ moles of $I_{2}$ are mixed and allowed to attain equilibrium at $50 0^{\circ}$ C. At equilibrium, the concentration of $H I$ is found to be $10$ moles. The equilibrium constant for the formation of $H I$ is

50

15

100

25

Equilibrium constant

$(K_{c}) = \frac{( H I ) ^{2}}{[ H _{2} ] [ I _{2} ]}$

$= \frac{10 \times 10}{10 \times 0.2}$

$= 50$

Q. 15 moles of H2​ and 5.2 moles of I2​ are mixed and allowed to attain equilibrium at 500∘C. At equilibrium, the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

50

15

100

25

Equilibrium constant (Kc​)=[H2​][I2​](HI)2​ =10×0.210×10​ =50

Q. $15$ moles of $H_{2}$ and $5.2$ moles of $I_{2}$ are mixed and allowed to attain equilibrium at $50 0^{\circ}$ C. At equilibrium, the concentration of $H I$ is found to be $10$ moles. The equilibrium constant for the formation of $H I$ is

Equilibrium constant

$(K_{c}) = \frac{( H I ) ^{2}}{[ H _{2} ] [ I _{2} ]}$

$= \frac{10 \times 10}{10 \times 0.2}$

$= 50$