### Chemical Equilibrium Problem#### For the reaction:\[ \text{2HOCl(g) } \leftrightarrow \text{ H}_2\text{O(g) } + \text{ Cl}_2\text{O(g) } \]Given that the value of the equilibrium constant (\(K\)) at 25°C is 11, calculate the value of the equilibrium constant in terms of partial pressure (\(K_p\)) for this reaction at 25°C.\[K_p = \]#### Explanation:1. **Chemical Equation:** The reaction shows the decomposition of hypochlorous acid (\(HOCl\)) gas into water (\(H_2O\)) and dichlorine monoxide (\(Cl_2O\)) gases.2. **Equilibrium Constant (K):** At 25°C, the equilibrium constant (\(K\)) is provided as 11. This value is dimensionless and represents the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants at equilibrium.3. **Calculate \(K_p\):** To find \(K_p\), the equilibrium constant in terms of partial pressures, you'll need to use the relationship between \(K\) and \(K_p\):\[ K_p = K(RT)^{\Delta n} \]Where:- \( R \) is the universal gas constant (0.0821 L·atm/mol·K)- \( T \) is the temperature in Kelvin- \( \Delta n \) is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).In this reaction:\[ \Delta n = (1 + 1) - 2 = 0 \]Therefore:\[ K_p = K \times (RT)^{0} \]\[ K_p = K \]Since \( \Delta n = 0 \), the \( K_p \) will be the same as \( K \).\[ K_p = 11 \]Please fill in the value in the box provided accordingly.

Answered: Image /qna-images/answer/6ce3acd2-0cc3-4d2d-b8cc-374840a33860.jpg

Answered: 2HOCI(g) # H,0(g) + Cl,0(g) the value…

Science

Chemistry

2HOCI(g) # H,0(g) + Cl,0(g) the value ofK at 25°C is II. Calculate the value of K, for this reaction at 25°C.

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

For the reaction CaO(s) + CO2(g) CaCO3(s), the value of AH is -178 kJ/mol. Using Le Chatelier's principle, which of the following changes to the system will shift the equilibrium to the right? i) Increasing the amount of CaO. ii) Decreasing the volume of the container. iii) Increasing the temperature. iv) Decreasing the temperature O A. only ii and iii OB. only i & ii OC. all of them OD. only i & iv OE. only i & iii
2NO(g) + 2H₂(g) = 2H₂O(g) + N₂(g) K = 0.110 AH° = -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' = [?] K' Enter
4. Sketch a graphical representation of the following events for the equilibrium below. PCIse 5 PCe + Cl ΔΗ = + 88 kJ/mol 15(g) 3(g) 2(g) a) PCI, is introduced into a flask at t=0 to a concentration of 2 mol/L. b) The flask is heated gently until no further changes in concentration of PCl, occur (5 minutes). At this time, the concentration of PCl(g) is 1.25 mol/L. c) At t=10 min an additional 0.25 mol/L of chlorine gas is injected into the flask. d) At t=15 min, no further changes in concentration were observed. Indicate the places on the graph where the system is at equilibrium. f) Use the information provided to calculate K, the equilibrium constant of this reactions. Conc'n (mol/L) Time (min)
Consider the following system at equilibrium where AH° = 92.7 kJ, and K. =1.80x104, at 298 K. NH,HS(s) NH3(g) + H2S(g) When some moles of NH HS(s) are added to the equilibrium system at constant temperature: The value of Ke The value of Qel |Ke The reaction must Orun in the forward direction to restablısh equilibrium. Orun in the reverse direction to restablish equilibrium O remain the same. It is already at equilibrium. The concentration of NH3 will| Submit Answer Retry Entire Group No more group attempts remain
7) The reaction: A + B = 2 C, can be represented by the equilibrium expression, Kc = [c]² [A][B] When 1.00 M of C was allowed to reach equilibrium and 0.062 M of A was formed. If this reaction was performed at the same temperature using 0.500 M C, what would the equilibrium concentration of A be? = 258 at 520K.
Need help understanding questions 6
4. A certain reaction has an equilibrium constant of 8500 at 298K. At 755K, K is 0.65. Calculate DHrxn- A) -50 kJ B) -40 kJ C) -30 kJ D) -20 kJ E) -10 kJ
4. The atmospheric oxidation of nitrogen monoxide, 2 NO (g) + O2 (g) = 2NO2 (g), was studied at 184°C with initial pressures of 1.00 atm of NO and 1.000 atm of O2. At equilibrium Po,- 0.56 atm. Calculate Kp.
Consider the reactions and their respective equilbrium constants: NO(g) + 1/2BR(g) > NOBR(g) KP = 5.3 2NO(g) > N2(g) + O2(g) KP = 2.1 x 1030 Use these reactions and their equilbrium constants to predict the equlibrium constant for the following reaction: N2(g) + O2(g) + Br2(g) > 2NOBR(g)
Heating HI(g) at 425 °C causes some of this compound to decompose, forming H2 (g) and I2 (g). Eventually, the amounts of the three species do not change further, the system has reached equilibrium. (At this point, approximately 22% of the HI has decomposed.) What is happening in this system at the molecular level? O H2(g) reacts with I2 (g) to produce HI(g) at the same rate as it decomposes. H2 (g) and I2 (g) molecules collide with HI(g) making the reaction stop. Only part of the HI(g) molecules has enough energy to decompose at this temperature. 5:55 PM FLV 2/22/2020 Type here to search hp ins prt sc delete home f12 f8 f9 f10 f5 f6 f7 14 10 num & backspace lock 23 8. 3 %24
Dry ice (solid carbon dioxide) is placed in a sealed container. It sublimes to produce carbon dioxide gas and reaches equilibrium at a given temperature. If the amount of dry ice in the container is doubled, then _____________. (After choosing the correct answer justify why in your own words why you chose that answer) CO2 (s) = CO2 (g) A)The amount of CO2 (g) would double. B)The amount of CO2 (g) would increase but not double. C)The amount of CO2 (g) would not change. D)The equilibrium amount of CO2 (s) would be less.
A sample of NO decomposes according to the following equation: 2 NO(g) 1 N₂(g) + 1 0₂(9) An equilibrium mixture in a 10-L vessel at 8529 °C, contains 0.00354 g of NO, 0.00283 g of N₂, and 0.00266 g of 0₂. (a) Calculate Kp for this reaction at this temperature. Kp = = (b) What is the total pressure exerted by the equilibrium mixture of gases? Ptotal = atm.