write the equilibrium constant expression for each reaction a.) NH3(g)+ CIF3(g)-><3HF(g) +1/2N(g) +1/2Cl2(g)

Question

write the equilibrium constant expression for each reaction a.) NH3(g)+ CIF3(g)-><3HF(g) +1/2N(g) +1/2Cl2(g)

Answer 1

You need to use whole number coefficents to do this.

2NH3(g)+ 2CIF3(g)-><6HF(g) + N2(g) + Cl2(g)

Keq=[HF]⁶[N2][Cl2]/[NH3]²[ClF3]²

Answer 2

To write the equilibrium constant expression for a reaction, we use the concentrations (or partial pressures for gaseous species) of the reactants and products involved in the reaction.

For the given reaction:
NH3(g) + CIF3(g) ⇌ 3HF(g) + 1/2N2(g) + 1/2Cl2(g)

The equilibrium constant expression can be written as:

Kc = ([HF]^3 * [N2]^0.5 * [Cl2]^0.5) / ([NH3] * [CIF3])

In this expression, the brackets [ ] represent the concentrations of the respective species involved in the reaction.

Note: The powers in the expression are determined by the stoichiometric coefficients of the balanced equation. If a species doesn't have a coefficient explicitly stated, its power is considered to be 1. In this case, for example, we have [N2] raised to the power of 0.5 because the coefficient is 0.5.