The equilibrium constant Kp 132 627 C for the reaction 2 S

The equilibrium constant, Kp = 1.32 @ 627 °C for the reaction: 2 SO3(g) --> 2 SO2(g) + O2(g)

What is the equilibrium constant (Kp ) for the reaction shown below at the same temperature?


SO2(g) + 1/2 O2(g) --> SO3(g)


Answer is 0.87 - just need steps - thanks!

Solution

since the constants have been halved and the reaction inverted, so the Kp will be square rooted and then taken the inverse.

so Kp=(1/1.32)^0.5

=0.87

 The equilibrium constant, Kp = 1.32 @ 627 °C for the reaction: 2 SO3(g) --> 2 SO2(g) + O2(g) What is the equilibrium constant (Kp ) for the reaction shown b

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