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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.

[tex] \sf \: SO_3 \leftrightharpoons \: SO_2 + \: \frac{1}{2} O_2[/tex]

[tex] \sf \: K_c = \dfrac{(SO₂) ( O₂) {}^{ \dfrac{1}{2} } }{SO_3} [/tex]

[tex] \sf \: 0.18= \dfrac{(SO₂) ( O₂) {}^{ \dfrac{1}{2} } }{SO_3} \: \: \: \: (1)[/tex]

Now, writing the Kc value of the reaction

SO₂ + ½ O₂ ⇌SO₃

[tex] \sf \: K_c' = \dfrac{SO_3 }{(SO₂) ( O₂) {}^{ \dfrac{1}{2} } } [/tex]

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

[tex] \sf \: K_c' = \dfrac{1}{K_c} [/tex]

[tex] \sf \: K_c' = \dfrac{1}{0.18} [/tex]

[tex] \sf \: K_c' = \dfrac{1}{ \bigg(\dfrac{18}{100}\bigg) } [/tex]

[tex] \sf \: K_c' = \dfrac{100}{ 18 } [/tex]

[tex] \sf \: K_c' = \dfrac{50}{ 9 } [/tex]

[tex] \sf \: K_c' = 5.55...[/tex]

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

L.C.M = 2×3×2×2×2×2×2×3×7

= 4032

Answer:

Hope it helps...

Explanation:

Solution: Corrosion can be prevented by creating a barrier between the object and oxygen. Galvanisation is a process to prevent corrosion and so is painting. However, exposing to sunlight is not a method to prevent corrosion.

Answer:

जैव विविधता जीवन और विविधता के संयोग से निर्मित शब्द है जो आम तौर पर पृथ्वी पर मौजूद जीवन की विविधता और परिवर्तनशीलता को संदर्भित करता है। संयुक्त राष्ट्र पर्यावरण कार्यक्रम (युएनईपी), के अनुसार जैवविविधता biodiversity विशिष्टतया अनुवांशिक, प्रजाति, तथा पारिस्थितिकि तंत्र के विविधता का स्तर मापता है।