**Chemical Equilibrium and Temperature Dependence**Enter your answer in the provided box.For reactions carried out under standard-state conditions, the equation ΔG = ΔH - TΔS becomes:\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]Assuming \( \Delta H^\circ \) and \( \Delta S^\circ \) are independent of temperature, one can derive the equation:\[ \ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]where \( K_1 \) and \( K_2 \) are the equilibrium constants at \( T_1 \) and \( T_2 \), respectively. Given that at 25.0°C, \( K_c \) is \( 4.63 \times 10^{-3} \) for the reaction:\[ N_2O_4 (g) \leftrightarrow 2NO_2 (g) \]and \( \Delta H^\circ = 58.0 \text{ kJ/mol} \),calculate the equilibrium constant at 40.0°C.**Equation to use:**\[ K_c = \boxed{} \]

This problem can be solved by using the given integrated Arrhenius equation as follows :

*Xog pəpIAojd For reactions carried out under standard-state conditions, the equation AG=AH- TAS becomes AG" = H - TAS". Assuming AH" and AS are independent of temperature, one can derive the equation: K2 AH T2-T1. In %3D K1 where K1 and K2 are the equilibrium constants at T, and T2, respectively. Given that at 25.0°C, K̟ is 4.63 x 103 for the reaction N,O,(g) 5 2NO,(g) AH° =58.0 kJ/mol calculate the equilibrium constant at 40.0°C. K =

*Xog pəpIAojd For reactions carried out under standard-state conditions, the equation AG=AH- TAS becomes AG" = H - TAS". Assuming AH" and AS are independent of temperature, one can derive the equation: K2 AH T2-T1. In %3D K1 where K1 and K2 are the equilibrium constants at T, and T2, respectively. Given that at 25.0°C, K̟ is 4.63 x 103 for the reaction N,O,(g) 5 2NO,(g) AH° =58.0 kJ/mol calculate the equilibrium constant at 40.0°C. 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

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
Consider the following equilibrium: 2NOC1 (g) 2NO(g) + Cl₂ (g) AG=41. KJ Now suppose a reaction vessel is filled with 3.76 atm of nitrosyl chloride (NOC1) and 6.26 atm of nitrogen monoxide (NO) at 1161. °C. Answer the following questions about this system: Under these conditions, will the pressure of NO tend to rise or fall? Is it possible to reverse this tendency by adding C1₂? In other words, if you said the pressure of NO will tend to rise, can that be changed to a tendency to fall by adding C1₂? Similarly, if you said the pressure of NO will tend to fall, can that be changed to a tendency to rise by adding Cl₂? If you said the tendency can be reversed in the second question, calculate the minimum pressure of C12 needed to reverse it. Round your answer to 2 significant digits. OO rise fall yes no atm x10 X Ś
Olawiany Tepical Use the References to access important values if needed for this question. Consider the following system at equilibrium where AH° = 108 kJ, and K. = 1.29×102 at 600 K: COC,(g) CO(g) + C(g) If the TEMPERATURE on the equilibrium system is suddenly decreased: The value of K. A. Increases B. Decreases C. Remains the same The value of Q. A. Is greater than K. B. Is equal to K. C. Is less than K. The reaction must: A. Run in the forward direction to restablish equilibrium. B. Run in the reverse direction to restablish equilibrium. C. Remain the same. Already at equilibrium. The concentration of Cl, will: A. Increase. B. Decrease. C. Remain the same. Submit Answer Retry Entire Group 9 more group attempts remaining Previ 50 hp
None
[Review Topics] [References] Use the References to access important values if needed for this question. Consider the following system at equilibrium where AH=-108 kJ, and K. 77.5, at 600 K. CO(g) + C,(g) COCI,(g) When 0.22 moles of Cl,(g) are added to the equilibrium system at constant temperature: The value of K The value of Qe K. The reaction must O run in the forward direction to restablish cquilibrium. O run in the reverse direction to restablish cquilibrium. O remain the same. It is already at cquilibrium. The concentration of CO will Submit Answer Retry Entire Group 1 more group attempt remaining
26) Suppose that at a pressure of 1 atm and temperature 300 K, a given material has two possible phases: liquid (L) and solid (S). We start the system with N₁ = 5 moles in the liquid phase and Ng = 9 moles in the solid phase. For these values of pressure and temperature, PL = 5 x 10-21 J • Hs9 x 10-21 J. What is the equilibrium value of NL? a. 9 mol b. 14 mol c. 1.49 mol d. 5 mol e. O mol
At -5.92 °C the pressure equilibrium constant K = 5.9 for a certain reaction. P Here are some facts about the reaction: • If the reaction is run at constant pressure, 127. kJ/mol of heat are released. • The constant pressure molar heat capacity C • The net change in moles of gases is -1. = 2.41 J mol -1. K¹. -1 р Yes. ☐ x10 Using these facts, can you calculate K at -27. °C? ○ No. If you said yes, then enter your answer at right. Round it to 2 significant digits. Р If you said no, can you at least decide whether K at -27. °C will be bigger or smaller than K, at −5.92 °C? П Yes, and K, will be р bigger. Yes, and K smaller. will be No. ☑
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.
[Review Topics] [References) Use the References to access important values if needed for this question. Consider the following system at equilibrium where AH° =-10.4 kJ/mol, and K. = 55.6 , at 698 K. H2 (g) + I2 (g)=2 HI (g) When 0.26 moles of HI (g) are added to the equilibrium system at constant temperature: The value of K. The value of Q. v Ke The reaction must Orun in the forward direction to restablish equilibrium. Orun in the reverse direction to restablish equilibrium. O remain the same. It is already at equilibrium. The concentration of I, will Submit Answer Retry Entire Group 9 more group attempts remaining (Previ ins prt sc delete fg f10 f12