**Chemical Equilibrium and Equilibrium Constant**In this educational module, we will explore the concept of chemical equilibrium and how to determine the equilibrium constant for a given reaction.### Original Reaction and Equilibrium DataConsider the following chemical reaction:\[ 2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons 2 \text{H}_2\text{O}(g) + \text{N}_2(g) \]The equilibrium constant (\( K \)) for this reaction is given as:\[ K = 0.110 \]Additionally, the enthalpy change (\( \Delta H^\circ \)) for this reaction is:\[ \Delta H^\circ = -707 \text{kJ} \]### Altered ReactionNow, we need to determine the equilibrium constant for an altered version of the original reaction:\[ \text{H}_2\text{O}(g) + \frac{1}{2} \text{N}_2(g) \rightleftharpoons \text{NO}(g) + \text{H}_2(g) \]Let's denote the equilibrium constant for this altered reaction as \( K' \).### Solution ApproachTo find \( K' \), consider the relationship between the original reaction and the altered reaction. Here, the altered reaction can be obtained by reversing and then halving the original reaction. When a reaction is reversed, the equilibrium constant for the reversed reaction is the reciprocal of the original equilibrium constant. Furthermore, when a reaction is halved, the equilibrium constant for the halved reaction is the square root of the original equilibrium constant.Therefore, the steps to calculate \( K' \) are as follows:1. Reverse the original reaction to get the reciprocal of \( K \).2. Halve the coefficients in the reversed reaction and take the square root of the resulting equilibrium constant.### Mathematical Calculation1. Reverse the original reaction:\[ 2 \text{H}_2\text{O}(g) + \text{N}_2(g) \rightleftharpoons 2 \text{NO}(g) + 2 \text{H}_2(g) \]\[ K_{\text{reversed}} = \frac{1}{K} = \frac{1}{0.110} \approx 9.09 \]2. Halve the reversed

Answered: Image /qna-images/answer/d5e4167d-2c5f-4dea-a554-18444a8222a6.jpg

Answered: 2NO(g) + 2H₂(g) = 2H₂O(g) + N₂(g) K =…

2NO(g) + 2H₂(g) = 2H₂O(g) + N₂(g) K = 0.110 ΔΗ° = −707 kJ What is the equilibrium constant for the reaction when it is altered as below? H₂O(g) + N₂(g) = NO(g) + H₂(g) 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

Calculate the value of Kp for the equation C(s) + CO₂(g) = 2 CO(g) given that at a certain temperature Kp = Kp = ? C(s) + 2 H₂O(g) = CO₂(g) + 2 H₂(g) H₂(g) + CO₂(g) = H₂O(g) + CO(g) Kpl = 3.93 = 0.769 Kp2 =
dont provide handwriting solution .....
The equilibrium constant for the following reaction is 20.9 at 907 °C. 2 SO2(g) + O2(g) 2 SO3(g) K= 20.9 at 907 °C Calculate the equilibrium constant for the following reactions at 907 °C. (a) 2 SO3(g) 2 SO2(g) + 02(g) K= (b) SO3(g) SO (g) + 1/2 O2(g) K=
The equilibrium constant for the reaction 2 N2(g) + 5 O2(g) ⇆ 2 N2O5(g) was measured at two temperatures: Temperature, K Kp 343 2.5x10−8 944 7.6x10−6 What is ΔHo for the reaction (in kJ/mol)?
Consider this system at equilibrium. A(aq) = B(aq) AH = -350 kJ/mol What can be said about Q and K immediately after an increase in temperature? Q > K because Q increased. = K because neither changed. Q > K because K decreased. Q < K because K increased. О<к because Q decreased. How will the system respond to a temperature increase? shift right shift left no change
Calculate AGO at 605 K for H₂O(g) + 1/2 02(g) → H₂O₂(g) using the following data: AGO = H₂(g) +O₂(g) → H₂O₂(g) K = 8.9 × 10³6 at 605 K 2H₂(g) + O₂(g) → 2H₂O(g) K = 2.0 x 106 at 605 K kJ/mol
Which of the following equilibrium expressions (K) represent the equilibrium reaction: 2 Fe(s) + 3 Cl2(g) = 2 FeCl3(s)
At -8.64 °C the concentration equilibrium constant K = 7.7 for a certain reaction. Here are some facts about the reaction: -1 • The initial rate of the reaction is 5.3 mol·L¯¹·s¯¹. . If the reaction is run at constant pressure, 132. kJ/mol of heat are absorbed. • The constant pressure molar heat capacity C = 2.97 J'mol K¹. Yes. Using these facts, can you calculate K at 16. °C? x10 O No. If you said yes, then enter your answer at right. Round it to 2 significant digits. If you said no, can you at least decide whether Kat 16. °C will be bigger or smaller than K at -8.64 °C? C Yes, and K will be bigger. Yes, and K will be smaller. No.
The equilibrium constant, Kc, is given for one of the reactions below. What is the value of the missing equilibrium constant, Ke? Cl₂(g) + H₂O(g) = 2 HCl(g) + O₂(g) K = 7.52 x 10-² 4 HCI(g) + O₂(g) = 2 Cl₂(g) + 2 H₂O(g) K = ? OK=177 K = 0.150 OK-5.66 x 10-3 OK 13.3 3 pts ) K = 3.65
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reaction. 2 NO (g) N2(g) + O2(g) Round your answer to 2 significant digits. K ☐ x10 ค
At 100° C, the Keq for the Haber process is 4,51 x 10-5: N2 + 3 H2 ó 2 NH3 Consider the following reaction conditions and determine if the system is at equilibrium. If not, indicate the direction in which the reaction must proceed to establish equilibrium pNH3 = 105 atm, pN2 = 35 atm and pH2 = 495 atm pNH3 = 35 atm, pN2 = 0 atm and pH2 = 595 atm pNH3 = 26 atm, pN2 = 202 atm and pH2 = 42 atm pNH3 = 105 atm, pN2 = 5 atm and pH2 = 55 atm