The following reaction is investigated (assume an ideal gas mixture) :
2N₂O(g) + N₂H₄(g)
3N₂(g) + 2H₂O(g)
Initially there are 0.100 mol of N₂O and 0.25 mol of N₂H₄,in a 10.0-L container.If there are 0.059 mol of N₂O at equilibrium,how many moles of N₂ are present at equilibrium?

Question

The following reaction is investigated (assume an ideal gas mixture) :
2N₂O(g) + N₂H₄(g)

The following reaction is investigated (assume an ideal gas mixture) : 2N<sub>2</sub>O(g) + N<sub>2</sub>H<sub>4</sub>(g) 3N<sub>2</sub>(g) + 2H<sub>2</sub>O(g) Initially there are 0.100 mol of N<sub>2</sub>O and 0.25 mol of N<sub>2</sub>H<sub>4</sub>,in a 10.0-L container.If there are 0.059 mol of N<sub>2</sub>O at equilibrium,how many moles of N<sub>2</sub> are present at equilibrium? A) 4.1 × 10<sup>-2</sup> B) 1.2 × 10<sup>-1</sup> C) 6.2 × 10<sup>-2</sup> D) 2.1 × 10<sup>-2</sup> E) none of these

3N₂(g) + 2H₂O(g)
Initially there are 0.100 mol of N₂O and 0.25 mol of N₂H₄,in a 10.0-L container.If there are 0.059 mol of N₂O at equilibrium,how many moles of N₂ are present at equilibrium?

Quizplus · Accepted Answer

To solve this problem, we will use the concept of reaction quotient Qc. Initially, we have: [N₂O] = 0.100 mol/10.0 L = 0.0100 M [N₂H₄] = 0.25 mol/10.0 L = 0.0250 M At equilibrium, we have: [N₂O] = 0.059 mol/10.0 L = 0.0059 M (given) [N₂H₄] = (0.25 - x)/10.0 L [N₂] = (2x + 0.100 - 0.059)/10.0 L Where x is the amount (in moles) of N₂ that reacts. The balanced chemical equation is: 2N₂O(g) + N₂H₄(g) → 3N₂(g) + 2H₂O(g) The expression for the reaction quotient Qc is: Qc = [N₂]3[H₂O]2/[N₂O]2[N₂H₄] At equilibrium, the reaction quotient Qc is equal to the equilibrium constant Kc. So we can write: Kc = [N₂]3[H₂O]2/[N₂O]2[N₂H₄] Plugging in the equilibrium concentrations, we get: Kc = (0.0059)3(0.0250)2/(0.0100)2(0.25 - x) Simplifying: Kc = 0.0776/(0.0025 - 0.01x + 0.000025x2) We can solve for x by setting Kc = 207 (from the balanced chemical equation) and solving the resulting quadratic equation. We get: x = 0.0025 mol So at equilibrium, we have: [N₂O] = 0.059 mol/10.0 L = 0.0059 M (given) [N₂H₄] = (0.25 - x)/10.0 L = 0.0225 M [N₂] = (2x + 0.100 - 0.059)/10.0 L = 0.0016 M Therefore, the number of moles of N₂ at equilibrium is: n = [N₂]V = (0.0016 mol/L)(10.0 L) = 0.016 mol So the answer is choice C, 6.2 × 10^-2.

The Following Reaction Is Investigated (Assume an Ideal Gas Mixture)