Equilibrium law and the expression for K

State, in terms of equilibrium concentrations, the expression for $K$, the equilibrium constant of this reaction.

By the time equilibrium was reached, the vessel held 1.7 mol of $\ce{H2(g)}$. Calculate the equilibrium amounts, in mol, of the three remaining gases, then use them to calculate $K$.

(i) An equilibrium mixture in a 25.0 $\mathrm{dm^3}$ reactor at 500 °C contains 15.0 mol $\ce{N2}$, 22.5 mol $\ce{H2}$ and 5.25 mol $\ce{NH3}$. Calculate the equilibrium constant, $K$, for the reaction. (ii) The volume of the reactor containing the equilibrium…

The following system is at equilibrium. $\ce{H2(g) + I2(g) <=> 2HI(g)}$ $K = 54$ What is the value of the equilibrium constant, K, for the reaction $\ce{2HI(g) <=> H2(g) + I2(g)}$? A. $-54$ B. $\dfrac{1}{54}$ C. $\sqrt{54}$ D. $\dfrac{1}{\sqrt{54}}$

(i) To check the value of $K$ found in (d)(viii), a student set up an experiment: 0.500 mol each of propan-1-ol and ethanoic acid were mixed with a small amount of concentrated sulfuric acid at 25 °C, and the mixture's volume was kept constant throughout.…

Nitrogen reacts with hydrogen to form ammonia. $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ A mixture of 1.50 mol of $\ce{N2(g)}$ and 4.20 mol of $\ce{H2(g)}$ was placed in a 1.00 $\mathrm{dm^3}$ container at a certain temperature. At equilibrium the mixture contained…

Butan-2-ol reacts with octadecanoic (stearic) acid to form an ester that is used as a lubricant. $\ce{C17H35COOH(l) + CH3CH(OH)CH2CH3(l) <=> C17H35COOCH(CH3)CH2CH3(l) + X(l)}$ (i) Identify the by-product X(l) and state the type of reaction taking place. (ii)…

Which expression correctly gives the equilibrium constant for the following reaction? $\ce{2NO(g) + Cl2(g) <=> 2NOCl(g)}$ A. $K_c = \dfrac{[\ce{NOCl}]^2}{[\ce{NO}]^2[\ce{Cl2}]}$ B. $K_c = \dfrac{[\ce{NO}]^2[\ce{Cl2}]}{[\ce{NOCl}]^2}$ C.…

The equation for a reaction between gases X and Y is: $\ce{2X(g) + Y(g) <=> 2Z(g)}$ At equilibrium at 700 K the concentrations of X, Y and Z are $0.50\ \mathrm{mol\,dm^{-3}}$, $0.40\ \mathrm{mol\,dm^{-3}}$ and $1.0\ \mathrm{mol\,dm^{-3}}$ respectively.…