### Equilibrium Concentration Calculation#### Problem Statement:Suppose a 250. mL flask is filled with 1.1 mol of \( NO_3 \) and 2.0 mol of \( NO \). The following reaction becomes possible:\[ NO_3 (g) + NO (g) \rightleftharpoons 2NO_2 (g) \]The equilibrium constant \( K \) for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of \( NO \). Round your answer to two decimal places.#### Given:- Volume of the flask = 250 mL (or 0.250 L)- Initial moles of \( NO_3 \) = 1.1 mol- Initial moles of \( NO \) = 2.0 mol- Equilibrium constant \( K = 8.88 \)#### Required:- Equilibrium molarity of \( NO \).#### Solution Approach:1. **Set up the ICE (Initial, Change, Equilibrium) table:** | Species | Initial (M) | Change (M) | Equilibrium (M) | |---------|-------------|------------|------------------| | \(NO_3\) | \( \frac{1.1}{0.250} \) | -x | \( \frac{1.1}{0.250} - x \) | | \(NO\) | \( \frac{2.0}{0.250} \) | -x | \( \frac{2.0}{0.250} - x \) | | \(NO_2\) | 0 | +2x | 2x |2. **Write the equilibrium expression for \( K \):** \[ K = \frac{[NO_2]^2}{[NO_3] \cdot [NO]} = 8.88 \]3. **Substitute equilibrium concentrations into the expression:** \[ K = \frac{(2x)^2}{\left( \frac{1.1}{0.250} - x \right) \left( \frac{2.0}{0.250} - x \right)} = 8.88 \]4. **Solve the equation for \( x \), which represents the change in concentration from the initial to

Given: Volume of flask = 250mL = 0.250L Initial moles of NO3 = 1.1 mol Initial moles of NO = 2.0 mol…

Suppose a 250. mL flask is filled with 1.1 mol of NO3 and 2.0 mol of NO. The following reaction becomes possible: NO3(g) + NO(g) → 2NO,(g) The equilibrium constant K for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

Suppose a 250. mL flask is filled with 1.1 mol of NO3 and 2.0 mol of NO. The following reaction becomes possible: NO3(g) + NO(g) → 2NO,(g) The equilibrium constant K for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Suppose a 500. mL flask is filled with 1.7 mol of Cl₂, 1.1 mol of CHCl3 and 1.3 mol of HCl. The following reaction becomes possible: Cl₂(g) + CHCl3 (g) → HC1(g) +CC14 (g) The equilibrium constant K for this reaction is 9.41 at the temperature of the flask. Calculate the equilibrium molarity of Cl₂. Round your answer to two decimal places. 10999
I am confused because I am getting two possible answers one being 10. 6 and the other being 1.38
Suppose a 250. mL flask is filled with 1.5 mol of CO, 0.80 mol of NO and 0.70 mol of CO,. The following reaction becomes possible: 2' NO, (g) +CO(g) - NO(g)+CO,(g) The equilibrium constant K for this reaction is 6.92 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. olo Ar OM ?
Suppose a 250. mL flask is filled with 0.70 mol of NO₂, 0.20 mol of NO and 1.9 mol of CO₂. The following reaction becomes possible: NO₂ (g) + CO (g) NO(g) + CO₂(g) The equilibrium constant K for this reaction is 0.715 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places.
229. Subject :- Chemistry
Suppose a 500.mL flask is filled with 1.5mol of CO, 1.1mol of H2O and 1.6mol of H2. The following reaction becomes possible: CO(g)+H2O(g)+CO2(g)+H2(g) The equilibrium constant K for this reaction is 0.840 at the temperature of the flask. Calculate the equilibrium molarity of CO2. Round your answer to two decimal places.
Suppose a 250. mL flask is filled with 1.1 mol of Cl2, 0.20 mol of CHCl3 and 2.0 mol of HCl. The following reaction becomes possible: Cl2(g) + CHCl3(g) HCl(g) +CCI4(g) The equilibrium constant K for this reaction is 0.652 at the temperature of the flask. Calculate the equilibrium molarity of HCI. Round your answer to two decimal places. Ом X G
Suppose a 500. mL flask is filled with 0.30 mol of SO, and 1.2 mol of SO3. This reaction becomes possible: 2SO, (g) +O,(g) 2SO3 Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O,. You can leave out the M symbol for molarity. so, SO2 O2 initial change equilibrium
Suppose a 250. mL flask is filled with 1.7 mol of NO2, 0.80 mol of CO and 0.70 mol of CO2. The following reaction becomes possible: NO2(g) + CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 2.90 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.
Suppose a 500. mL flask is filled with 1.5 mol of H₂ and 1.2 mol of Cl₂. The following reaction becomes possible: H₂(g) + Cl₂(g) → 2HCl(g) The equilibrium constant K for this reaction is 3.45 at the temperature of the flask. Calculate the equilibrium molarity of H₂. Round your answer to two decimal places. M Ś
Suppose a 250. mL flask is filled with 0.30 mol of NO and 0.80 mol of NO2. The following reaction becomes possible: NO3(g)+NO(g)<->2NO2(g) The equilibrium constant K for this reaction is 0.717 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places.
Do