**Chemical Equilibrium and Equilibrium Constant Changes****Original Reaction:**\[ \text{SO}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons \text{SO}_2\text{Cl}_2(\text{g}) \]\[ \Delta H^\circ = -264 \, \text{kJ} \quad K = 34.5 \]**Altered Reaction:**\[ 3\text{SO}_2(\text{g}) + 3\text{Cl}_2(\text{g}) \rightleftharpoons 3\text{SO}_2\text{Cl}_2(\text{g}) \]\[ K' = [ \, ? \, ] \]---**Problem Statement:**When the reaction is altered as shown above, what is the value of the new equilibrium constant?*Hint: Consider how the equilibrium constant \( K \) changes when the coefficients of the balanced equation are multiplied by a factor.*---**Explanation:**- The problem involves determining the new equilibrium constant \( K' \) for the altered reaction.- The alteration involves multiplying all coefficients in the original reaction by 3.In equilibrium constant calculations, if the coefficients of a balanced chemical equation are multiplied by a factor \( n \), the new equilibrium constant \( K' \) is given by:\[ K' = K^n \]Thus, for this specific case:\[ K' = (34.5)^3 \]**Interactive Section:**___\[ \boxed{\begin{array}{c}\text{Enter the value of } K' \\\text{} \\\text{K'} \quad \text{[ \, \, \, \, \, \, \, \, ]} \quad \text{Enter}\end{array}} \]---

Answered: Image /qna-images/answer/d8052a27-d026-47f7-b7d8-e565bb63b5ec.jpg

SO₂(g) + Cl₂(g) = SO₂Cl₂(g) ΔΗ° = −264 kJ K = 34.5 When the reaction is altered, as below, what is the value of the new equilibrium constant? 3SO₂(g) + 3Cl₂(g) = 3SO₂Cl₂(g) K' = [?]

SO₂(g) + Cl₂(g) = SO₂Cl₂(g) ΔΗ° = −264 kJ K = 34.5 When the reaction is altered, as below, what is the value of the new equilibrium constant? 3SO₂(g) + 3Cl₂(g) = 3SO₂Cl₂(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

6. Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction: 2H2S(g) + SO2(9) =Ss(s, rhombic) + 2H20(1) What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
The equilibrium constant, Kp, for the following reaction is 2.74 at 1150 K.2SO3(g) 2SO2(g) + O2(g)If ΔH° for this reaction is 198 kJ, what is the value of Kp at 1030 K?Kp = --- The equilibrium constant, Kp, for the following reaction is 2.01 at 500 K.PCl3(g) + Cl2(g) PCl5(g)If ΔH° for this reaction is -87.9 kJ, what is the value of Kp at 416 K?Kp =
12. For the following reaction at equilibrium in a closed container, ΔΗ-128 kJ 2 NaHCO3 (s) = NażCO3 (s) + H2O (g) + CO2 (g) State the effect (increase, decrease, or no change) of decreasing the temperature on the number of moles of sodium carbonate, Na2CO3. Note that Na2CO3 is a solid (this is a heterogeneous equilibrium); its concentration will remain constant, but its amount can change.
The Keg for the equilibrium below is 7.52 · 102 at 480.0 °C. 2Cl2 (g) + 2H2O (g) = 4HCI (g) + O2 (g) What is the value of Keg at this temperature for the following reaction? HCI (g) + 1/202 (g) = 1/2C12 (g) + 1/2H2O (g) 1.91 7.52. 10-2 1.88x10-2 0.524
Consider the synthesis of ammonia: N2(g) + 3H2(g)<=>2NH3(g) At 926.5 °C, the equilibrium constant Kp is 4.09 × 10–5. What is the value of Kc?
Cl3] ² [Hz] ² [A] [AC] Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25 °C: N₂ (g) + 3 H₂ (g) = 2 NH3 (g) K = 5.6 x 105 Calculate the equilibrium constant for the following reaction at 25 °C: NH3 (g) 1/2 N₂ (g) + 3/2 H₂ (g) K' = ? Kc = [NH3 ]² [₂] [H₂]] Kc- [~₂] ¹/2 [H NH3
SO₂ reacts with Cl₂ to form SO₂Cl₂. The equilibrium constant, Kp, for this reaction is 0.417 at 373 K. 4 SO₂(g) + Cl₂(g) =SO₂Cl₂(g) The standard enthalpy change for this reaction (AH°) is -67.2 kJ/mol. (a) Predict the effect on the production of SO₂Cl₂ when the temperature of the equilibrium system is increased. (b) Use the van't Hoff equation to estimate the equilibrium constant for this reaction at 449 K. Kat 449 K =
At 13.7 °C the concentration equilibrium constant K = 2.0 × 10 for a certain reaction. Here are some facts about the reaction: • C The constant pressure molar heat capacity C₁ = 2.49 J-mol K. •If the reaction is run at constant pressure, 147. kJ/mol of heat are released. •If the reaction is run at constant pressure, the volume increases by 13.%. ○ Yes. Using these facts, can you calculate K, at 3.8 °C? x10 No. If you said yes, then enter your answer at right. Round it to 2 significant digits. n If you said no, can you at least decide whether Kat 3.8 °C will be bigger or smaller than K at -13.7 °C? Yes, and K will be bigger. Yes, and K will be smaller. No. x
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)?
The following reaction takes place into a constant-volume flask at 427°C: CCL(g) = C(s) +2 C2(g) AH=132 kJ At 427°C, the equilibrium constant for the reaction given above is K. = 1.3×10-2. A flask is filled with 2.00 atm of CCl4 at 427°C. (a) What is the value of K for this reaction? (R=0.082 L.atm/mol.K) (b) What are the partial pressures of CClL, Ch and the total pressure at equilibrium? (c) How will the equilibrium pressure of CCl(g) change if the temperature is increased? (d) How will the equilibrium system respond if the volume of the flask is reduced?
Consider the following equilibrium: 2NO, (g) – N,04(g) AG = - 5.4 kJ Now suppose a reaction vessel is filled with 7.26 atm of dinitrogen tetroxide (N,O) at 95. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of N,O, tend to rise or fall? 4 fall Is it possible to reverse this tendency by adding NO,? In other words, if you said the pressure of N,0, will tend to rise, can that yes be changed to a tendency to fall by adding N0,? Similarly, if you said the no pressure of N,0, will tend to fall, can that be changed to a tendency to rise by adding NO,? If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO, needed to reverse it. atm Round your answer to 2 significant digits. O O