1.2 moles of SO3 are allowed to dissociate in a 2 litre vessel the reaction is 2SO3(g)⇔2SO2(g)+O2(g) and the concent ration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be

initial : 1.2 0 0 2SO3 ⇌ 2SO2+ O2 at ⇌1.2 -2x 2x x at equilibrium concentration of O2 = 0.1 nO22=0.1 nO2=0.2=xTotal =1.2-2x + 2x + x =1.2+0.2 =1.4

1.2 moles of SO3 are allowed to dissociate in a 2 litre vessel the reaction is 2SO3(g)⇔2SO2(g)+O2(g) and the concent ration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be

initial : 1.2 0 0 2SO3 ⇌ 2SO2+ O2 at ⇌1.2 -2x 2x x at equilibrium concentration of O2 = 0.1 nO22=0.1 nO2=0.2=xTotal =1.2-2x + 2x + x =1.2+0.2 =1.4

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1.2 moles of SO₃ are allowed to dissociate in a 2 litre vessel the reaction is $2 {SO}_{3 (g)} \Leftrightarrow 2 {SO}_{2 (g)} + O_{2 (g)}$ and the concent ration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be

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$initial : 1.2 0 0 2 {SO}_{3} ⇌ 2 {SO}_{2} + O_{2} at ⇌ 1.2 - 2 x 2 x x at equilibrium concentration of O_{2} = 0.1 \frac{n_{O_{2}}}{2} = 0.1 n_{O_{2}} = 0.2 = x$

Total =1.2-2x + 2x + x =1.2+0.2 =1.4

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1.2 moles of SO3 are allowed to dissociate in a 2 litre vessel the reaction is 2SO3(g)⇔2SO2(g)+O2(g) and the concent ration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be

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1.2 moles of SO₃ are allowed to dissociate in a 2 litre vessel the reaction is $2 {SO}_{3 (g)} \Leftrightarrow 2 {SO}_{2 (g)} + O_{2 (g)}$ and the concent ration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be