Question #286322

Equilibrium Constant

Q1

for both of the following equilibria, write the expression for the equilibrium constant Kc and state it’s units:

i) H2(g) + I2 (g) —-> <—— 2HI (g) (2 marks)

ii) 2SO2 (g) + O2(g) ——> <—— 2SO2 (g) (2 marks)

Q2

the reaction for the formation of hydrogen bromine does not go to completion but reaches an equilibrium, as shown:

H2(g) + Br2 —-> <—— 2HBr(g)

a mixture of 1.9 mol of H2 and 1.9 of Br was prepared and allowed to reach equilibrium in a closed vessel on 250 cm3 capacity. The resulting equilibrium mixture was found to contain 3.0 moles of HBr. Calculate the value of K. (3 marks please detailed answers with explanation🙏🏾🙏🏾🙏🏾🙏🏾)


1
Expert's answer
2022-01-18T18:11:02-0500

Q1

(i) kc=[HI]2[H2][I2]k_c=\frac{[HI]^2}{[H_{2}][I_{2}]}


Units Mole L-1


(ii) kc=[SO2]2[SO2]2[O2]k_c = \frac{[SO_2]^2}{[SO_2]^2[O_2]}


Units Mole-1L


Q2


H2+Br22HBrI1.91.90Cxx2xE1.9x1.9x2x\begin{matrix} & H_2+ &Br_2&\rightleftharpoons&2HBr \\ I&1.9&1.9& & 0\\ C&x&x&&2x\\ E&1.9-x&1.9-x&&2x \end{matrix}



But 2x=3    x=1.52x=3\implies\>x=1.5

[H2]=1.91.5[H_2]=1.9-1 .5

=0.4mol/250cm3=0.4mol/250cm^3

=1.6mol/L=1.6mol/L


[Br2]=1.6mol/L[Br_2]=1.6mol/L

[HBr]=3×1000250=12mol/L[HBr]=\frac{3×1000}{250}=12mol/L


Kc=[HBr]2[H2][Br2]=1221.6×1.6=56.25K_c=\frac{[HBr]^2}{[H_2][Br_2]}=\frac{12^2}{1.6×1.6}=56.25







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