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 hydrogen iodide. 2HIG) P H2(g) +12(g) [H2] [12] K=— =1.98x10-2 at 895 K (HI) A flask originally contains 0.373 M hydrogen iodide. Calculate the equilibrium concentrations of the three gases. [HI] = [H2) = [12] =
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] =
6) Distrubuting Equilibrium a) The equilibrium constant, K, for the following reaction is 10.5 at 350 K 2CH2Cl2(g)CH4(g + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.32x10-2 M CH2C2, 0.172 M CH4 and 0.172 M CCI4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 9.58x102 mol of CH4(g is added to the flask? CH2Cl2l [CH4 [CCI4] b) The equilibrium constant, K, for the...
1) 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 0.311 M HI, 4.18×10-2 M H2 and 4.18×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.85×10-2 mol of I2(g) is added to the flask? 2) The equilibrium constant, K, for the following reaction is 1.20×10-2 at...
The equilibrium constant, K, for the following reaction is 5.10x10-6 at 548 K. NH4CI(s)H3(g +HCI(g) An equilibrium mixture of the solid and the two gases in a 1.00 L flask at 548 K contains 0.200 mol NH4CI, 2.26x10-3M NH3 and 2.26x103 M HCI. If the concentration of NH3(g) is suddenly increased to 3.76x10-3 M, what will be the concentrations of the two gases once equilibrium has been reestablished? [NH3] [HCI] The equilibrium constant, K, for the following reaction is 10.5...
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)
The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) CH4(g) + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.39×10-2 M CH2Cl2, 0.175 M CH4 and 0.175 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.118 mol of CCl4(g) is added to the flask?
Consider the following equilibrium system at 821 K. 2NOCI(g) 2NO(g) + Cl2 (g) If an equilibrium mixture of the three gases at 821 K contains 9.34 x 103 M NOCI 3.47 x 102 M NO, and 2.57 x 10 M Cl2, what is the value of the equilibrium constant К? К We were unable to transcribe this imageConsider the following equilibrium system at 521 K 2CH2 Cl2 (g) CH4 (g)CC4 (g) If an equilibrium mixture of the three gases at...
Determining Equilibrium Concentrations and Constants Consider the following reaction: 2 NO(8)N2 (8)+0 (8) K,= 0.145 If a mixture of 1.15 M NO(g), 0.560 M N2 (g) and 0.560 M 02 (g) is placed in a reaction flask, calculate the equilibrium concentration of Na 2. Consider the following reaction: 2 N2 (g) +O (82NO (g) A 5.00 L flask was filled with 0.500 atm N20(g) and 0.500 atm He(g) at 500.0 °C. At equilibrium, the pressure of Ox(g) is found to...
The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) CH4(g) + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.09E-2 M CH2Cl2, 0.165 M CH4 and 0.165 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.82E-2 mol of CH2Cl2(g) is added to the flask? [CH2Cl2] = M [CH4] = M [CCl4] = M