A mixture of 1.397 g of H2 and 70.76 g of Br2 is heated in a...

A mixture of 1.397 g of H2 and 70.76 g of Br2 is heated in a 2.00 L vessel at 700. K. These substances react according to H2(g) + Br2(g) ↔ 2 HBr(g). At equilibrium the vessel is found to contain 0.592 g of H2. (a) Calculate the equilibrium concentrations of H2, Br2, and HBr. (b) Calculate Kc.

Expert Solution

H2 moles = 1.397 / 2 = 0.6985

H2 molarity = moles / volume = 0.6985 / 2 = 0.349 M

Br2 molarity = (70.76 / 159.8 ) x 1/2 = 0.221 M

equilibrium H2 molarity = ( 0.592 / 2 ) x 1/2 = 0.148 M

H2(g) + Br2(g) <---------------------------> 2 HBr(g).

0.349 0.221 0 ------------------> initial

-x -x +2x --------------------> changed

0.349-x 0.221-x 2x ----------------------> equilibrium

but at equilibrium H2 molarity = 0.148 M

0.349 -x = 0.148

x = 0.201

now equilibrium concentrations

[H2] = 0.148 M

[Br2] = 0.221-x = 0.221 -0.201= 0.02 M

[HBr ] = 2x = 2 x 0.201 = 0.402 M

equilibrium constant Kc = [HBr]^2 / [H2][Br2]

Kc = (0.402 )^2 / (0.02)(0.148)

Kc = 54.6

queen_honey_blossom answered 1 year ago

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