Use K and initial concentrations to calculate equilibrium concentrations. Consider the equilibrium system involving the decomposition...
Use K and initial concentrations to calculate equilibrium concentrations. Consider the equilibrium system involving the decomposition of iodine monobromide. 21Br(g) = [Bry] [12] Bry(g) +12(g) Ke -8.22x10-3 at 462 K [IBr] A flask originally contains 0.261 M iodine monobromide. Calculate the equilibrium concentrations of the three gases. [IBr] =( [Bry] = [12] = TEM M M
Use K and initial concentrations to calculate equilibrium concentrations. Consider the equilibrium system involving the decomposition of nitrogen monoxide. 2NO(g) N2(g) + O2(g) [N2] [02] K=— = 3.62x10-2 at 286 K [NO] A flask originally contains 0.239 M nitrogen monoxide. Calculate the equilibrium concentrations of the three gases. [NO] = [N2] = [02] =
Calculate the equilibrium concentrations of H2, I2, and HI at 700 K if the initial concentrations are [H2] = 0.200 M and [I2] = 0.400 M. The equilibrium constant Kc for the reaction following reaction is 57.0 at 700 K. (Show Work) H2(g)+I2(g)<--- ---->2HI(g)
Consider the following chemical reaction: If the equilibrium concentrations of H,, I,, and HI are 0.15M, 0.033M, and 0.55M respectively. The value of K for this reaction is rte Gold catalyzes the decomposition of hydrogen iodide into its elements according to the equation: 2HI (g) 2(g) + H2(g). Consider the illustration at the right. What type of catalyst would this be: heterogeneous or homogeneous? ive a reason for your choice. Consider the following chemical reaction: If the equilibrium concentrations of...
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
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) goes to H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.308 M HI, 4.14×10-2 M H2 and 4.14×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.67×10-2 mol of I2(g) is added to the flask? [HI] = __M [H2] = __M [I2] = __M