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:
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
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