Question

A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700 K . These substances react as follows:

H2(g)+Br2(g)⇌2HBr(g)

At equilibrium the vessel is found to contain 0.566 g of H2.

1A) Calculate the equilibrium concentration of H2.

1B) Calculate the equilibrium concentration of Br2.

1C) Calculate the equilibrium concentration of HBr.

1D) Calculate Kc.

Answer 1

Answer:

Explanation:

Kc is defined as the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature

Step 1: Write the balanced the chemical equation

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

Step 2: Calculate the Concentrations by dividing the given moles by volume of container.

Initial H₂ = (1.374 g / 2.016 g/mole ) = 0.687 mole = ( 0.687 moles / 2 L ) = 0.341 M

Initial Br₂ = (70.31 g / 159.808 g/mole ) = 0.4399654586 mole = ( 0.44 moles / 2 L ) = 0.22 M

Equilibrium H₂ = (0.566 g / 2 g/moles ) = 0.287 mole = ( 0.287 moles / 2 L ) = 0.140 M

Step 3: Write the ICE table

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

Initial amount 0.341 (0.22)   0

Change amount -x -x +2x

Equilibrium amount 0.341-x 0.22-x 2x

Since the value of H₂ at equlibrium we have calculated [ H₂] = 0.140 M = [0.341-x]

hence value of x = 0.201 M

Hence all value at equilibrium is

1.C [HBr] = 2x = (2 × 0.201 ) M = 0.402 M

1.B [Br₂] = 0.22-x = (0.22-0.202) = 0.019 M

1.A [H₂] = 0.341-x = (0.3435-0.202) = 0.140 M

1.D Step 4: Calculate the Kc

Kc = [HBr]² / × [H₂] × [Br₂] = ( 0.402)² / ( 0.140 ) ×(0.019) = 60.75