Use the References to access important values if needed for this question.The equilibrium constant, K, for the following reaction is 1.80×10-² at 698 K.2HI(g)H2(g) +12(g)An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.302 M HI,4.05×102 M H2 and 4.05×102 M 12. What will be the concentrations of the three gasesonce equilibrium has been reestablished, if 2.13×10-2 mol of H2(g) is added to the flask?[HI][H2]=MM[12] =M

To solve this problem, we need to use the given equilibrium constant K=1.80×10−2 and the initial…

Answered: Use the References to access important…

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...

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The equilibrium constant, K., for the following reaction is 5.10×10-6 at 548 K. NH,C(s)NH3(g) + HC(g) Calculate the equilibrium concentration of HCl when 0.501 moles of NH,CI(s) are introduced into a 1.00 L vessel at 548 K. [HCI] = |M
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The equilibrium constant, Kc, for the following reaction is 1.80 × 10-4 at 298 K. NH4HS(s) NH3(g) + H₂S(g) If an equilibrium mixture of the three compounds in a 4.30 L container at 298 K contains 3.12 mol of NH4HS(s) and 0.310 mol of NH3, the number of moles of H₂S present is | mol.
The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500K.PCl5(g) PCl3(g) + Cl2(g)If an equilibrium mixture of the three gases in a 13.6 L container at 500K contains 0.230 mol of PCl5(g) and 0.374 mol of PCl3, the equilibrium concentration of Cl2 is M.
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H₂(g) + 1₂(g) ? An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.330 M HI, 4.43×10-² M H₂ and 4.43×10-2 M 1₂. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.69x10-2 mol of H₂(g) is added to the flask? [HI] = [H₂] = [1₂] = ΣΣΣ
-2 The equilibrium constant K,, for the following reaction is 7.73×10atmat 350°C. What is the value of K. for the reaction at the same temperature? N2(g) +3H2(g)→ 2NH3(g)
CO2 and H, are allowed to react until an equilibrium is established as follows: CO2 (g) + H, (g)H,0 (g) + CO (g) What will be the effect on the equilibriumn of removing CO from the equilibrium mixture?
The equilibrium constant, Ke, for the following reaction is 7.00 x 10-5 at 673 K. to access important values if needed for this question. NH₂I(s) ⇒ NH3(g) + HI(g) If an equilibrium mixture of the three compounds in a 4.33 L container at 673 K contains 1.71 mol of NH₂I(s) and 0.445 mol of NH3, the number of moles of HI present is mol.
A.) The equilibrium constant, K, for the following reaction is 2.90×10-2 at 1.15×103 K.2SO3(g) 2SO2(g) + O2(g)If an equilibrium mixture of the three gases in a 16.3 L container at 1.15×103 K contains 0.431 mol of SO3(g) and 0.309 mol of SO2, the equilibrium concentration of O2 is ______________________ M. ( fill in the blank ) B.) The equilibrium constant, K, for the following reaction is 6.50×10-3 at 298 K.2NOBr(g) 2NO(g) + Br2(g)If an equilibrium mixture of the three gases in a 19.7 L container at 298 K contains 0.433 mol of NOBr(g) and 0.500 mol of NO, the equilibrium concentration of Br2 is_______________________ M. ( fill in the blank)
References Use the References to access important values if needed for this question. A student ran the following reaction in the laboratory at 688 K: 2HI(g) H2(g) + I½(g) When he introduced HI(g) at a pressure of 4.87 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of I2(g) to be 0.510 atm. Calculate the equilibrium constant, K,, he obtained for this reaction. %3D Submit Answer
HBr=.....M
For which one of the following reactions will the equilibrium mixture contain mostly products? (a) 2 Cu (s) + O2 (g) → 2 CuO (s), Kc = 4 x 1045 (b) H+ (aq) + CIO2- (aq) + HCIO2 (aq), K. = 9.1 x 101 (c) 2HB1 (g) + H2 (g) + Br2 (g), Kc = 2 x 10-19 (d) All of the above.

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