For the reaction 2SO-2- - O-2-rightleftharpoons 2SO-3- the unit of equilibrium constant is-L mol-1-mol L-1-J mol-1-L-2 mol-1-

Question

Toppr Admin · Accepted Answer

Equilibrium constant of the given reaction can be written as-2SO2-g-O2-g-x21CB-2SO3-g-Unit of concentration -mole-xA0-L-x2212-1-x21D2-Keq-SO3-2-SO2-2-O2-xA0-L-mol-x2212-1Hence- option A is correct-

For the reaction $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ , the unit of equilibrium constant is:
$L m o l^{- 1}$
$m o l L^{- 1}$
$J m o l^{- 1}$
$L^{2} m o l^{- 1}$

Equilibrium constant of the given reaction can be written as:

$2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g)$

Unit of concentration $= m o l e L^{- 1}$

$\Rightarrow K_{e q} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} L . m o l^{- 1}$

Hence, option A is correct.

For the reaction 2SO2+O2⇌2SO3, the unit of equilibrium constant is:L mol−1mol L−1J mol−1L2 mol−1

Equilibrium constant of the given reaction can be written as:2SO2(g)+O2(g)⇋2SO3(g)Unit of concentration =mole L−1⇒Keq=[SO3]2[SO2]2[O2] L.mol−1Hence, option A is correct.

For the reaction $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ , the unit of equilibrium constant is:
$L m o l^{- 1}$
$m o l L^{- 1}$
$J m o l^{- 1}$
$L^{2} m o l^{- 1}$

Equilibrium constant of the given reaction can be written as:

$2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g)$

Unit of concentration $= m o l e L^{- 1}$

$\Rightarrow K_{e q} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} L . m o l^{- 1}$

Hence, option A is correct.