### Calculating the Equilibrium Constant#### Problem Statement:Calculate the equilibrium constant for the chemical reaction, given the equilibrium concentrations of the reactants and products in the table.The reaction is:\[ 2 \text{AB(aq)} + 4 \text{CD(aq)} \rightleftharpoons 2 \text{AD(aq)} + 2 \text{BD(aq)} + 4 \text{C(aq)} \]#### Equilibrium Concentrations:| Substance | Equilibrium concentration ||-----------|-----------------------------|| AB | 4.99 M || CD | 2.54 M || AD | 4.02 M || BD | 2.28 M || C | 1.27 M |#### Formula for Equilibrium Constant, \( K \):For the reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is given by:\[ K = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]#### Applying the Equilibrium Concentrations:For the given reaction:\[ 2 \text{AB(aq)} + 4 \text{CD(aq)} \rightleftharpoons 2 \text{AD(aq)} + 2 \text{BD(aq)} + 4 \text{C(aq)} \]The equilibrium constant \( K \) is:\[ K = \frac{[\text{AD}]^2 [\text{BD}]^2 [\text{C}]^4}{[\text{AB}]^2 [\text{CD}]^4} \]Substitute the given equilibrium concentrations:\[ K = \frac{(4.02)^2 (2.28)^2 (1.27)^4}{(4.99)^2 (2.54)^4} \]Calculate \( K \):\[ K = \frac{(16.1604) (5.1984) (2.6013)}{(24.9001) (41.6061)} \]\[ K = \frac{(16.1604 \times 5.1984 \times 2.6013)}{(

The equilibrium constant is the ratio of the products of concentrations of products to that of the…

Calculate the equilibrium constant for the chemical reaction, given the Substance Equilibrium concentration equilibrium concentrations of the reactants and products in the table. AB 4.99 M CD 2.54 M 2 AB(aq) + 4 CD(aq) = 2 AD(aq) + 2 BD(aq) + 4 C(aq) AD 4.02 M BD 2.28 M 1.27 M K =

Calculate the equilibrium constant for the chemical reaction, given the Substance Equilibrium concentration equilibrium concentrations of the reactants and products in the table. AB 4.99 M CD 2.54 M 2 AB(aq) + 4 CD(aq) = 2 AD(aq) + 2 BD(aq) + 4 C(aq) AD 4.02 M BD 2.28 M 1.27 M K =

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

@quinnipiac.edu doesn't allow editing on a Mac. To learn more, contact your admin about your Office plan. 4. What is the correct equilibrium constant (K.) expression for the following reaction? Fe2O3(s) + 3H2(g) 2 2Fe(s) + 3H20(g)
Here are the equilibrium constants for two reactions of nitrogen with oxygen. What is the equilibrium constant for N₂O (g) + 1/2O₂(g) → 2NO (g)? N₂(g) + O₂(g) = 2NO(g) K= 4.1 × 10^-31 N₂(g) + 1/2O₂(g) = N₂O(g) K=2.4 × 10^-18
Consider this reaction equation: A2B(g)= 2 A(g) + B(g) If you start with 1.2 M A2B, and if at equilibrium, 50% of it has decomposed, what are the equilibrium concentrations of A and B? What is the value of the equilibrium constant?
For the reaction below, Kc =9.2 x 10 5. What is the equilibrium concentration of D if the reaction begins with 0.64 M A? A (aq) + 2 B (s)=C (s) + 2 D (aq) tv
For the chemical equilibrium N₂ + 3H22NH3 what is the expression for the equilibrium constant? K=
Given the equilibrium equation: 2A(g) + B2(g) 2AB(g) Solve for the molar concentration of A when Keq = 0.0719, [B2] = 3.15 M and [AB] 0.895 M %3D О 39.2 М 1.88 M О 0.25 М 3.54 М
Predict which species will be oxidized and which will be reduced in a molten salt mixture that contains CdI2, NaOH, and AlCI3. Cd2+(aq) + 2e → Cd(s) Eored = – 0.40 V I2(s) + 2e → 2 I–(aq) Eored = + 0.54 V Na+(aq) + 1e → Na(s) Eored = – 2.71 V Cl2(g) + 2e → 2 Cl–(aq) Eored = + 1.36 V Al3+(aq) + 3e → Al(s) Eored = – 1.66 V O2(g) + 2 H2O(l) + 4e → 4 OH–(aq) Eored = + 0.40 V which is oxidized which is reduced
Using the above equilibrium expressions, calculate the equilibrium constant, K, for the following concentrations: 3. [HCl] = 1.2 x 10-3 M, [O2] = 3.8 x 10-4 M, [H2O] = 5.8 x 10-2 M, [Cl2] = 5.8 x 10-2 M The equilibrium equation: Keq = [Cl2]^2 [H2O]^2 / [HCl]^+ [O2] ) [NH3] = 0.62 M, [H2] = 0.14 M, [N2] = 0.45 M. the equilibrium equation: Keq= [NH3]^-2 / [H2]^3 [N2] [H2] = 0.0045 M, [Cl2] = 0.0045 M, [HCl] = 0.0625 M. the equilibrium equation: Keq= [HCl]^2 / [H2] [Cl 2] [SO2] = 2.00 M, [O2] = 1.50 M, [SO3] = 3.00 M. the equilibrium equation: Keq = [so3]^2/ [so2]^2 [O2] Please answer all the questions to the best of your ability.
image solutions step by step
The reaction of bromine gas with chlorine gas, shown below, has a Kc of 7.20 at 200 oC. If a closed vessel was charged with the two reactants, each at an initial concentration of 0.50 M, what would be the equilibrium concentration of Cl2(g)?
Most reaction do not go to completions. Equilibrium is established between the reactants and products. For the reaction of A with B to produce C and D, we can write the chemical equation as: аА + bв + сс + dD where + represents equilibrium between the reactants and products. The extent to which the reaction proceeds to product formation, at a given temperature, is given by the equilibrium constant, Kę. The equilibrium constant is written mathematically (for the above equation) as: [C[[D]* K. = [A]°[B]" where, a, b, c, and d are the stoichiometric coefficients from the balanced chemical equation and the brackets, [ ], indicate molar concentration. 1. For the following reaction: 2A + В + 3с calculate the equilibrium constant, K, if at equilibrium the concentration of A is 0.15M, the concentration of B is 0.20M, and the concentration of C is 0.10M.
CH4(g) + 2 HS(g) = CS, (g) + 4 Ha(g) A reaction mixture initially contains 0.50 M CH, and 0.75 M HS. If the equilibrium concentration of H, is 0.44 M, find the equilibrium constant (K) for the reaction.