-8203-at a certain temperature the concentration of H2-g-I2-g- AND HI IN AN EQUILIBRIUM MIXTURE WERE 0-2g-L nbsp-7-62g-L and 19-2g-L respectively- Calculate the equilibrium constant Kc for the reaction H2-g-I2-g-EQUILIBRIUM 2HI-g- -8203-

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Byju's Answer

Standard XII

Chemistry

Potassium Dichromate

#8203;at a ce...

Question

at a certain temperature the concentration of H2,(g)I2(g) AND HI IN AN EQUILIBRIUM MIXTURE WERE 0.2g+L ,7.62g+L and 19.2g+L respectively. Calculate the equilibrium constant Kc for the reaction H2(g)+I2(g)EQUILIBRIUM 2HI(g)

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Solution

Dear Student,

$H_{2} (g) + I_{2} (g) \leftrightarrow 2 HI (g) At eq . 0.2 7.62 19.2 K_{C} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} or, K_{C} = \frac{(19.2)^{2}}{(0.2) (7.62)} = \frac{368.64}{1.524} = 241.889 = 2.418 \times 10^{2}$

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​at a certain temperature the concentration of H2,(g)I2(g) AND HI IN AN EQUILIBRIUM MIXTURE WERE 0.2g+L ,7.62g+L and 19.2g+L respectively. Calculate the equilibrium constant Kc for the reaction H2(g)+I2(g)EQUILIBRIUM 2HI(g)​

Dear Student, H2(g) + I2(g) ↔ 2HI(g)At eq. 0.2 7.62 19.2 KC=[HI]2[H2][I2]or, KC=(19.2)2(0.2)(7.62)=368.641.524=241.889 = 2.418×102

at a certain temperature the concentration of H2,(g)I2(g) AND HI IN AN EQUILIBRIUM MIXTURE WERE 0.2g+L ,7.62g+L and 19.2g+L respectively. Calculate the equilibrium constant Kc for the reaction H2(g)+I2(g)EQUILIBRIUM 2HI(g)

Dear Student,

$H_{2} (g) + I_{2} (g) \leftrightarrow 2 HI (g) At eq . 0.2 7.62 19.2 K_{C} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} or, K_{C} = \frac{(19.2)^{2}}{(0.2) (7.62)} = \frac{368.64}{1.524} = 241.889 = 2.418 \times 10^{2}$