Question 2 The AGʻrxn for the reaction: A (g) + B (g) 2C is +192 kJ/mol at 25 °C. What is the equilibrium constant? 9.72 x 10-6 0.986 2.61 x 10-2 2.30х 10:34
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- Question 4 Determine the equilibrium constant for the following reaction at 382 °C. HCN (g) + 2 H2 (g) ⇌ CH3NH2 (g) ΔH° = –158 kJ; ΔS° = –219.9 J/K ANSWER:13.0 Just need Part 2 Is there a temperature at which equilibrium for the reaction in Question 4 will "jump the arrows?" Why or why not? If you determined that temperature will shift the position of the equilibrium for this reaction, at what temperature will that occur?QUESTION 6 The value of the equilibrium constant for the following reaction is 24 at a given temperature: N2(g) + 3H2(g) ⇌ 2NH3(g) If [N2] = 10.0 M, [H2] = 10.0 M, and [NH3] = 10.0 M, determine if the reaction is at equilibrium. If it is not, in which direction will it proceed to reach equilibrium? Since Q = 0.01, the reaction is not at equilibrium, it will shift to the left. It is not possible to tell if the reaction is at equilibrium. Since Q = 1, the reaction is not at equilibrium, it will shift to the left. Since Q = 1, the reaction is not at equilibrium, it will shift to the right. Since Q = 100, the reaction is not at equilibrium, it will shift to the right. Since Q = 100, the reaction is not at equilibrium, it will shift to the left. The reaction is at equilibrium. To reach equilibrium, the value of Keq must change. Since Q = 0.01, the reaction is not at equilibrium, it will shift…Consider the hypothetical reaction, which has an equilibrium constant (Kp) of 16.1 when the reaction occurs at 167°C. 2 W (g) + X (g) <---> 3 Y (g) If the equilibrium partial pressure of W is 0.787 atm and that of Y is 1.65 atm, what will the equilibrium partial pressure (in atm) of X be? (answer in 3 significant figures)
- Given the following two measurements of the equilibrium constant for a reaction, calculate ΔH° for the reaction. (kJ/mol) (no scientific notation) T, °C K 30.0 5.2 x 103 50.0 15.4 x 104There are two questions here if that is okay...Question 1 Consider the reaction: 2 A (aq) ⇌ B(aq) Given the following KC values and starting with the initial concentration of A = 4.00 M, complete ICE diagram(s)and find the equilibrium concentrations for A and B.A) KC = 4.00B) KC = 200C) KC = 8.00 x10-3 Question 2 Consider the reaction: Cl2 (g) + F2 (g) ⟷ 2 ClF (g) KP=? The partial pressure of 203 kPa for Cl2 and a partial pressure of 405 kPa for F2. Upon reaching equilibrium, thepartial pressure of ClF is 180 kPa. Calculate the equilibrium concentrations and then find the value for KP.issue 6Choose the equilibrium in which the products are favored by a drop in pressure and the reactants favored by a drop in temperature.a) N2(g) + 3 H2(g) ⇔ 2 NH3(g) + 92.3 kJb) H2(g) + I2(g) + 51.8 kJ ⇔ 2 HI(g)c) PCl3(g) + Cl2(g) ⇔ PCl5(g) ΔH = -84.2 kJd) 2 NO2(g) ⇔ 2 NO(g) + O2(g) ΔH = +54 kJ
- Determine the equilibrium constant for the following reaction at 25 deg. C: 2I- (aq) + Br2(l) --> I2(s) + 2Br-(aq). (Example of answer: 6.5 x 10e-12, where 12 is exponent)Consider the following equilibrium: N2O4(g) ⇄2NO2(g). If Kp = 0.144 at 950.0 K, calculate Kc. (no scientific notation)Given: O2 (g) + 2HF (g) -> <- OF2 (g) + H2O (g) ∆H = + 318 kJ What will happen if the temperature of the system is increased? A. Equilibrium will shift towards left B. Equilibrium will shift towards right C. No shift D. No effect
- Solve all part please. Consider the equilibrium reaction N2O4 (g) ⇌ 2 NO2 (g)a. Using appendix C, find H and S for this reaction (you will need this for b,c). delta s= 176.6J delta H=5802kJb. Knowing that K is temperature dependent and that G = -RTlnK, find the equilibrium constant (K) and T whenthe above reaction is at equilibrium with PNO2 = PN2O4 = 1.0 atm.c. Knowing that K is temperature dependent and that G = -RTlnK, find the equilibrium constant (K) and T whenthe above reaction is at equilibrium with PT = 10.0 atm and PNO2 = 2 PN2O4.In the reaction 450.0⁰c: caco3(s) ⇌ cao(s) + co2(g), at equilibrium, if pco2 = 0.0160 atm , what is the value of KcCalculate T (in K) given ΔG^0= -19.4 kJ/mol and ΔH^0= 69.9 kJ/mol and ΔS^0= 253 J/mol KΔG^0=ΔH^0−TΔS^0 A) 706 B) 1,760 C) 353 D) 0.353 E) 118