Question

Use the information below to answer questions 33-35. Hydrogen iodide decomposes according to the equation: 2HI(g) ^ H2(g)+h_2(g). The equilibrium constant K_e is 0.0156 at 400 c. Suppose a 0.660 mol sample of HI was infected into a 2.00 L reation vessel held at 400 c 33. Calculate the equilibrium concentration of H_2. A.0.053 M B. 0.023 M C.0.013 M D.0.073 M E.0.033 M 34. Calculate the equilibrium concentration of l_2. A.0.053 M B. 0.023 M C. 0.013 M D0.073 M E. 0.033 M 35. Calculate the equilbrium concentration of HI A. 0.264 M b. 0.224 M C.0.284 M D. 0.204 M E.0.244 M

Answer 1

we know that

concentration = moles / volume (L)

so

initially

[HI] = 0.66 / 2 = 0.33

now

2HI ----> H2 + I2

using ICE table

at equilibrium

[HI] = 0.33 -2x

[H2] = x

[I2] = x

now

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

so

0.0156 = [x] [x] / [0.33-2x]^2

0.0156 = [x]^2 / [0.33-2x]^2

solving we get

x = 0.033

so

33)

at equilibrium

[H2]= x = 0.033 M

option E) 0.033 M

34)

[I2] = x = 0.033 M

option E) 0.033 M

35)

[HI] = 0.33 - 2x

[HI] = 0.33 - ( 2 * 0.033)

[HI] = 0.264 M

option A) 0.264 M