The following reaction occurred when a 1.0-liter reaction vessel was initially charged with 2.0 moles of N₂(g) and 4.0 moles of H₂(g) : 3H₂(g) + N₂(g) 2NH₃(g)
Once equilibrium was established,the concentration of NH₃(g) was determined to be 0.57 M at 700.°C.The value for K_c at 700.°C for the formation of ammonia is:

Question

The following reaction occurred when a 1.0-liter reaction vessel was initially charged with 2.0 moles of N₂(g) and 4.0 moles of H₂(g) : 3H₂(g) + N₂(g)

The following reaction occurred when a 1.0-liter reaction vessel was initially charged with 2.0 moles of N<sub>2</sub>(g) and 4.0 moles of H<sub>2</sub>(g) : 3H<sub>2</sub>(g) + N<sub>2</sub>(g) 2NH<sub>3</sub>(g) Once equilibrium was established,the concentration of NH<sub>3</sub>(g) was determined to be 0.57 M at 700.°C.The value for K<sub>c</sub> at 700.°C for the formation of ammonia is: A) 3.2 × 10<sup>-1</sup> B) 6.1 × 10<sup>-3</sup> C) 1.1 × 10<sup>-1</sup> D) 6.0 × 10<sup>-2</sup> E) none of these

2NH₃(g)
Once equilibrium was established,the concentration of NH₃(g) was determined to be 0.57 M at 700.°C.The value for K_c at 700.°C for the formation of ammonia is:

Quizplus · Accepted Answer

The balanced equation is: 3H₂(g) + N₂(g) 2NH₃(g) Initially, we have: [H₂] = (4.0 mol) / (1.0 L) = 4.0 M [N₂] = (2.0 mol) / (1.0 L) = 2.0 M [NH₃] = 0 M At equilibrium, let x be the molar concentration of NH3 formed. Then the equilibrium concentrations of the reactants and product can be expressed in terms of x: [H₂] = (4.0 - 3x) M [N₂] = (2.0 - x) M [NH₃] = x M The equilibrium constant expression is: K_c = ([NH₃]^2 / ([H₂]^3[N₂])) Substituting the equilibrium concentrations and solving for x gives: K_c = (x^2 / [(4.0 - 3x)^3(2.0 - x)]) 6.1 × 10^-3 = x^2 / [(4.0 - 3x)^3(2.0 - x)] 6.1 × 10^-3(4.0 - 3x)^3(2.0 - x) = x^2 From here, we could use the quadratic equation to solve for x, but it is simpler to use an iterative method to solve for x. By trial and error or using a numerical method such as Newton-Raphson, we find that x = 0.57 M is the solution that satisfies the equation. Therefore, the equilibrium constant is: K_c = (0.57^2 / [(4.0 - 3(0.57))^3(2.0 - 0.57)]) K_c = 6.1 × 10^-3 Therefore, the best choice is B.

The Following Reaction Occurred When a 1