State the effect of the following on the rection $2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)} + 189.4$ kJ at equilibrium:
(i) Temperature is increased.
(ii) Concentration of $S O_{2}$ is increased.
(iii) Pressure is decreased.
(iv) Helium is added at constant presuure.

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Solution

(i) Equilibrium will shift in backward direction which means backward reaction will be favoured.
(ii) Equilibrium will shift in forward direction or more of $S O_{3}$ will be formed.
(iii) Backward reaction will be favoured.
(iv) Backward reaction will be favoured.

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2 S O_{3} (g) ⇌ 2 S O_{2} (g)

S O_{2}

S O_{3}

S O_{2}

α

P_{2}

K_{p}

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g), Δ H = - X

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

[S O_{3}] = 1.90 M

_{c}

2 S O_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g) + X c a l

{S O}_{2}

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