Want to check work ### Chemical Equilibrium Problem**Consider the following system at equilibrium where ΔH° = -198 kJ, and Kₒ = 34.5, at 1150 K.**\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]When **0.29 moles of O₂(g)** are **removed** from the equilibrium system at constant temperature:1. **The value of Kₒ:** - **remains the same.** (Dropdown Selection)2. **The value of Qₒ:** - **is greater than** Kₒ. (Dropdown Selection)3. **The reaction must:** - **run in the reverse direction to reestablish equilibrium.** (Radio Button Selection)4. **The concentration of SO₂ will:** - **decrease.** (Dropdown Selection)#### Explanation:1. **Equilibrium Constant (Kᶜ):** - Kᶜ is a constant for a given temperature, so removing O₂ does not change the value of Kᶜ. It remains the same.2. **Reaction Quotient (Qᶜ):** - The reaction quotient Qᶜ is used to determine the direction the reaction must proceed to return to equilibrium. When O₂ is removed, Qᶜ becomes greater than Kᶜ, indicating that the system is no longer at equilibrium.3. **Direction to Reestablish Equilibrium:** - When Qᶜ > Kᶜ, the reaction must shift in the reverse direction (towards reactants) to reestablish equilibrium.4. **Concentration of SO₂:** - To counter the removal of O₂ and shift the reaction towards the reactants, the concentration of SO₂ will decrease as the equilibrium reestablishes itself.

Answer:- This question is answered by using the simple concept of change in the equilibrium when the…

Consider the following system at equilibrium where AH° = -198 kJ, and K. = 34.5, at 1150 K. %3D %3D 250-(g) + O2(g) 2503(g) When 0.29 moles of O2(g) are removed from the equilibrium system at constant temperature: The value of K. (remains the same. : The value of Q[ is greater than JKc- The reaction must run in the forward direction to restablish equilibrium. o run in the reverse direction to restablish equilibrium. oremain the same. It is already at equilibrium. The concentration of SO2 will ( decrease.

Consider the following system at equilibrium where AH° = -198 kJ, and K. = 34.5, at 1150 K. %3D %3D 250-(g) + O2(g) 2503(g) When 0.29 moles of O2(g) are removed from the equilibrium system at constant temperature: The value of K. (remains the same. : The value of Q[ is greater than JKc- The reaction must run in the forward direction to restablish equilibrium. o run in the reverse direction to restablish equilibrium. oremain the same. It is already at equilibrium. The concentration of SO2 will ( decrease.

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

Question 1 of 20 A barrel of crude oil has a volume of 42 gallons, only approximately 45% of which is processed into gasoline. If your car achieves 32 mi/gal, and you drive 36,000 miles in one year, how many barrels of crude oil are required to run your car for a year? Tap here or pull up for additional resources
Need help with question #24 Results Trial 1 : Initial Volume = 0 0.5143g=KHP 14.3 = Buret Measurment Trial 2: Initial Volume= 0 0.5615g = KHP 14.1 = Buret Measurement
What is the specific gravity of a substance that displaces 1.2 mL and has a mass of 5.021 Select one: O a. 4.2 b. 4.2 g/ mL С. 4.184 O d. 4.184 g/ml Oe. 4.18
2. You plot Ink vs 1/T and the trendline is y = -8702x+51.09. a. Calculate AH. b. Calculate AS. c. Calculate AG°.
5. This question has two parts a. and b. a. Assuming 100% yield, how much hexynyl lithium (in mmol) is produced from the reaction of LiHMDS with 1- hexyne? (mmol) b. Insert an image (.jpeg, png, .heic) showing all of your calculations for the moles of hexynyl lithium above. Note: Click on the image icon in the box below to upload/attach an image file. W
Please show work Thank you!
RESET Cancel Done Convert 5.890×10³ kg to ug:
AutoSave We hw OFF ... Home Insert Draw Design Layout References Mailings Review View O Tell me R Share O Comments Calibri (Bo.. v 11 v A A E vE v E v E E Aa v AaBbCcDdEe AaBbCcD AaBbCcDdE AaBb AaBbCcDdEe BIU V ab A • I v A v Paste х, х E= == No Spacing Normal Heading 1 Heading 2 Title Styles Pane Dictate Sensitivity C13 NMR Peaks Aldehydes RCO)R Aldehydes and ketones Carbaxylic R(CO)X Carbaxylic acid derivatives Ntrile Nitrile RCN CC C-C Alkyne Akyne R-CC-R RCH20 RCH2-O R4C RAC R3CH R3CH RCH2X X= C-C, C-O, Br, CI, N RCH2X R2CH2 RECH2 RCH3 RCH3 TMS TMS 220 200 180 160 140 120 100 80 60 40 20 Typical chemical shifts in 13c-NMR 9) Why do the peaks of associated with aldehydes (both in H NMR and C NMR) appear so much further downfield than other peaks? C13 NMR Peaks E Page 5 of 5 E English (United States) O Focus 295 words 白