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Physical Chemistry Please help 4) Consider the following reaction at 298.15 K: H2(g) +12(g) = 2HI(g)...
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) ----> H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.306 M HI, 4.10×10-2 M H2 and 4.10×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.208 mol of HI(g) is added to the flask? [HI] = ______ M [H2] = ______ M [I2] = ______M
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.319 M HI, 4.27×10-2 M H2 & 4.27×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.224 mol of HI(g) is added to the flask? [HI] = M [H2] = M [I2] = M please help me!
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) ⇌H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.320 M HI, 4.29×10-2 M H2 and 4.29×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.233 mol of HI(g) is added to the flask? [HI] = ___M [H2] = ___ M [I2] = ___M
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.302 M HI, 4.05×10-2 M H2 and 4.05×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.203 mol of HI(g) is added to the flask?
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.329 M HI, 4.41×10-2 M H2 and 4.41×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.54×10-2 mol of H2(g) is added to the flask? [HI] = M [H2] = M [I2] = M
Consider the following reaction: 2HI(g) =H2(g) +12(9) If 1.87 moles of HI, 0.333 moles of H2, and 0.277 moles of Iare at equilibrium in a 14.7L container at 888 K, the value of the equilibrium constant, Kp. is Submit Answer Retry Entire Group 9 more group attempts remaining
Consider the reaction: H2(g) +12(9) =2HI(9) A reaction mixture in a 3.60 -L flask at 500 K initially contains 0.375 g H2 and 17.91 g 12. At equilibrium, the flask contains 17.76 g HI. Part A Calculate the equilibrium constant at this temperature. Express your answer using three significant figures. IVO AXDA O O ? K = Submit Previous Answers Request Answer X Incorrect; Try Again
(1). The equilibrium constant, Kp, for the following reaction is 1.80×10-2 at 698K. 2HI(g) =H2(g) + I2(g) If an equilibrium mixture of the three gases in a 15.5 L container at 698K contains HI at a pressure of 0.399 atm and H2 at a pressure of 0.562 atm, the equilibrium partial pressure of I2 is atm. (2). Consider the following reaction: PCl5(g) =PCl3(g) + Cl2(g) If 1.17×10-3 moles of PCl5, 0.217 moles of PCl3, and 0.351 moles of Cl2 are at...
Consider the chemical reaction below at a given temperature and at equilibrium: H2(g) +12(g) = 2HI(g) Kc = 53.3 If 0.800 mol of H2 and 0.800 mol of 12 are placed in a 1.00L container and allowed to react, what is the [HI] when the reaction reaches equilibrium? [HIN In the expression for K N- (H2] [12]' The equilibrium concentrations can be expressed as follows: NOTE: This is NOT asking for the concentrations you solve for this is literally asking...
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.316 M HI, 4.24×10-2 M H2 and 4.24×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.21×10-2 mol of I2(g) is added to the flask?