Express the equilibrium constant for the following reaction.

7 N2(g) + 21 H2(g) ⇔ 14 NH3(g)

Question

Express the equilibrium constant for the following reaction.

7 N2(g) + 21 H2(g) ⇔ 14 NH3(g)

BlackTom AI · Accepted Answer

The problem asks for the equilibrium constant expression for the reaction: 7 N2(g) + 21 H2(g) ⇔ 14 NH3(g).
First, recall the general rule: for a balanced gaseous reaction aA + bB ⇌ cC, the equilibrium constant expression (Kc) is given by Kc = [C]^c / ([A]^a [B]^b), using concentrations in mol/L at constant temperature.
In this reaction, the products are NH3 with coefficient 14, and the reactants are N2 with coefficient 7 and H2 with coefficient 21.
Applying the rule, the numerator contains the product concentration raised to its stoichiometric power, and the denominator contains each reactant concentration raised to its stoichiometric power.
Thus, the correct Kc expression is: Kc = [NH3]^{14} / ([N2]^{7} [H2]^{21}).
If one were to use partial pressures instead of concentrations (Kp), the expression would be Kp = (P_NH3)^{14} / ((P_N2)^{7} (P_H2)^{21}), provided the temperature is the same and appropriate unit conventions are applied.
Note about the provided options: the list of answer choices appears to be missing in this prompt, so there are no specific choices to evaluate for correctness here. If options were given, you would compare each to the derived expression, checking that the exponents match the balanced equation and that NH3 appears in the numerator while N2 and H2 appear in the denominator with their respective stoichiometric powers.
In summary, the central idea is assigning NH3 to the numerator with exponent 14, and placing N2 and H2 in the denominator with exponents 7 and 21 respectively to reflect the balanced equation.

Express the equilibrium constant for the following reaction. 7 N2(g) + 21 H2(g) ⇔ 14 NH3(g)

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