Question

3. The equilibrium constant for the reaction SO3 (g) ⇌ SO2 (g) + ½ O2 (g) is 0.18 at 900 K. What will be the equilibrium constant for the reaction SO2 (g) + ½ O2 (g) ⇌ SO3 (g)?
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Answer 1

Exρlαnαtion:

Here they have been given that equilibrium constant of reaction SO₃ ⇌ SO₂ + ½ O₂ is 0.18 at temperature of 900K . Here they have been asked to calculate the value of equilibrium constant for the reaction of SO₂ + ½ O₂ ⇌SO₃. Let's see the process.

$\sf \: SO_3 \leftrightharpoons \: SO_2 + \: \frac{1}{2} O_2$

$\sf \: K_c = \dfrac{(SO₂) ( O₂) {}^{ \dfrac{1}{2} } }{SO_3}$

$\sf \: 0.18= \dfrac{(SO₂) ( O₂) {}^{ \dfrac{1}{2} } }{SO_3} \: \: \: \: (1)$

Now, writing the Kc value of the reaction

SO₂ + ½ O₂ ⇌SO₃

$\sf \: K_c' = \dfrac{SO_3 }{(SO₂) ( O₂) {}^{ \dfrac{1}{2} } }$

By observing Kc and Kc` we can say that kc = 1/kc`

$\sf \: K_c' = \dfrac{1}{K_c}$

$\sf \: K_c' = \dfrac{1}{0.18}$

$\sf \: K_c' = \dfrac{1}{ \bigg(\dfrac{18}{100}\bigg) }$

$\sf \: K_c' = \dfrac{100}{ 18 }$

$\sf \: K_c' = \dfrac{50}{ 9 }$

$\sf \: K_c' = 5.55...$

The equilibrium constant for the reaction SO2 (g) + ½ O2 (g) ⇌ SO3 (g) is 5.55..