
i need this in less than 20 minutes please if possible!!! For the following reaction, 2HI(g)...
Consider the reaction: H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.69 −L flask at 500 K initially contains 0.377 g H2 and 17.93 g I2. At equilibrium, the flask contains 17.72 g HI. Calculate the equilibrium constant at this temperature.
For the reaction in the previous problem, that is, 2HI(g) ↔ H2(g) + I2(g) Keq = 0.016 Initially a container contains 0.39 M HI and no product. What is the equilibrium concentration of H2?
For the reaction: 2HI(g) ↔ H2(g) + I2(g) Keq = 0.016 Initially a container contains 0.60 M HI, 0.038 M H2, and 0.15 M I2 at equilibrium. What is the new equilibrium concentration of H2, if the H2 concentration is increased by 0.276 M?
The equilibrium constant for the following reaction is 1.80x102 at 698K 2HI(8)H2( ) +I2(g) If an equilibrium mixture of the three gases at 698K contains 2.35x102 M HI(g) and 2.63x10-2 M H, what is the equilibrium concentration of I? Submit Answer Retry Entire Group 9 more group attempts remaining
Consider this reaction: H 2 (g)+ I 2 (g)⇌2HI(g) H2(g)+I2(g)⇌2HI(g) A rxn mixture in a 3.73 L flask at a certain temperature initially contains 0.766 g H2 and 96.7 g I 2 I2 . At equilibrium, the flask holds 90.3 g HI . Calculate the equilibrium constant (Kc)(Kc) for the reaction at this temperature. Record your answer using two significant figures.
Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.70 L flask at a certain temperature initially contains 0.766 g H2 and 97.0 g I2 At equilibrium, the flask contains 90.2 g HI. Calculate the equilibrium constant (Kc) for the reaction at this temperature. I keep getting 700 as the answer but it's not right.
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 0.306 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 0.320 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 0.329 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