CALCULATION:We know that, KP = Kc(RT)∆ngKP and Kc are equilibrium constant.R is GAS constantT is temperature∆ng is DIFFERENCES of moles.∴ If ∆ng ≠ 0 then Kp ≠ KcNow, 2C (s) + O2 (g) ⇌ 2CO (g)∆ng = +1∴ KP = Kc (RT)1Hence, Kp ≠ Kc "> CALCULATION:We know that, KP = Kc(RT)∆ngKP and Kc are equilibrium constant.R is GAS constantT is temperature∆ng is DIFFERENCES of moles.∴ If ∆ng ≠ 0 then Kp ≠ KcNow, 2C (s) + O2 (g) ⇌ 2CO (g)∆ng = +1∴ KP = Kc (RT)1Hence, Kp ≠ Kc ">
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In which one of the following equilibria, Kp ≠ Kc?

Current Affairs Chemistry in Current Affairs . 5 months ago

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CALCULATION:We know that, KP = Kc(RT)∆ngKP and Kc are equilibrium constant.R is GAS constantT is temperature∆ng is DIFFERENCES of moles.∴ If ∆ng ≠ 0 then Kp ≠ KcNow, 2C (s) + O2 (g) ⇌ 2CO (g)∆ng = +1∴ KP = Kc (RT)1Hence, Kp ≠ Kc
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Posted on 22 Nov 2024, this text provides information on Current Affairs related to Chemistry in Current Affairs. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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