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For N2(g) + 3H2(g) forming 2NH3(g), how does Kp relate to Kc?

AKp = Kc x (RT)^2, since delta n is plus two
BKp = Kc / (RT)^2, since delta n is minus two
CKp = Kc x (RT), since delta n is plus one here
DKp = Kc exactly, since delta n is zero for this
Answer & Solution
Correct answer: B. Kp = Kc / (RT)^2, since delta n is minus two
1. The conversion rule is Kp = Kc times (RT) raised to the power delta n. 2. Delta n is the moles of gaseous product minus the moles of gaseous reactant. 3. Gaseous product here is 2 mol of ammonia, so the product total is 2. 4. Gaseous reactants are 1 mol of nitrogen plus 3 mol of hydrogen, giving 4. 5. Delta n = 2 minus 4 = minus 2. 6. So Kp = Kc times (RT) to the power minus two, which is Kc divided by (RT) squared. 7. Reading delta n as plus two reverses the count and inverts the factor. 8. Delta n is zero only when the two sides carry equal moles of gas, which they do not here. _Source: OpenStax Chemistry (CC BY 4.0), Ch 13 "Fundamental Equilibrium Concepts", section 13.2 Equilibrium Constants_
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