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3 years ago

14. As the temperature of the reaction is increased,

Chemistry

1 answer:

AfilCa [17]3 years ago

3 0

Reactant molecules collide more frequently and with greater energy per collision

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Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init

Jlenok [28]

Answer : The initial concentration of HI and concentration of $HI$ at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of $H_2$ and $I_2$ = 0.11 M

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of $K_c$ will be,

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}$

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of $HI$ at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

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