A5.00 L reaction vessel at 445 degree C initially contained 0.100 moles each of

Question

A5.00 L reaction vessel at 445 degree C initially contained 0.100 moles each of H_2 and I_2 gases. These reacted to form $11 as described by the following equation. H_2(g) + I_2(g) doubleheadarrow 2HI(g) Kc = 50.0 at 445 degree C How many moles of HI were present in the reaction vessel when the reaction reached equilibrium? a) 1.85 times 10^-5 mol b) 4.21 times 10^-4 mol c) 3.84 times 10^-5 mol d) 5.13 times 10^-2 mol e) 1.56 times 10^-1 mol If a solution has pH of 3.82, what is the hydronium ion concentration ([H_2O])? a)3.6 times 10^-5 M b) 9.7 times 10^-5 M c) 1.5 times 10^-4 M d) 3.2 times 10^4 M e) 6.4 times 10^4 M

Explanation / Answer

(14) Initial concentration = 0.100 / 50.0 =0.00200 M

H2 (g) + I2 (g) < --------- > 2 HI (g) , Kc = 50.0

0.00200 0.002000            0.000 M           at initial

0.00200-x   0.00200-x       2x                     at equilibrium

Kc = [HI]2 / [H2][I2]

50.0 = (2x)2 / (0.00200-x)(0.00200-x)

50.0 ( 0.00000400 - 0.00400x + x2) = 4x2

46x2 - 0.200x + 0.000200 = 0

23 x2 - 0.100 x + 0.000100 = 0

Applying quadratic equation to solve x,

x = [-(-0.100) +/- sqrt.((-0.100)2 - (4 * 23 * 0.000100))] / (2*23)

x = 0.00156 M

Thereofre, [HI] = 2 * x = 2 * 0.00156 = 0.00312 M

Moles of HI = 0.00312 * 5.00 = 1.56 * 10-2mol

(e)

(15)

pH = 3.82

- Log[H3O+] = 3.82

[H3O+] = 10-3.82

[H3O+] = 1.5 * 10-4 M

(c)

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