Determine the rate law and the value of k for the following reaction using the data provided.
8. Using the data provided in the table, a) write the rate law and b) determine the value of the rate constant k for the following reaction: CO(g)+ CI2(g)COC12(g) [COl, (M) C2l, (M) Initial Rate (M/s) 0.25 0.50 0.25 0.40 0.80 0.80 0.696 3.94 1.97
What concentration of CO should be used in trial 3 in order for the researcher to get enough information to determine its reaction order? CO(g)+ Cl2(g) COCI2g) Initial Rate [COli (M) Clli (M)(M-15-1) Trial 1 0.25 Trial 2 0.50 Trial 3 0.40 0.696 0.80 3.94 0.80 1.97 0.80 M 0.25 M O 0.40 M O 0.50 M
Determine the rate law and the value of k for the following reaction using the data provided. 2 N2O5(g) → 4 NO2(g) + O2(g) [ N2O5]i (M) Initial Rate (M-1s-1) 0.033 1.84 × 10-4 0.066 7.37 × 10-4 0.099 1.66× 10-3 A. Rate = 5.6 × 10-1 M-1/2s-1[N2O5]3/2 B. Rate = 1.7 × 10-1 M-1s-1[N2O5]2 C. Rate = 1.7 × 10-1M1/2s-1[N2O5]1/2 D. Rate = 6.0 × 10-1 M-2s-1[N2O5]3 E. Rate = 5.2 × 10⁻1 s-1[N2O5]
Determine the rate law and the value of k for the following reaction using the data provided. 2 NO(g) + Br2(g) → 2 NOBr(g) [NO]i (M) [Br2]i (M) Initial Rate (Ms-1) 0.030 0.0055 1.81 x 10-2 0.030 0.0110 3.62 x 10-2 0.060 0.0055 7.25 x 10-2 please show step by step.
The equilibrium constant, K, for the following reaction is 1.29E-2 at 600 K. COCl2(g) <--> CO(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.297 M COCl2, 6.19E-2 M CO and 6.19E-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.13E-2 mol of CO(g) is added to the flask? [COCl2] = M [CO] = M [Cl2] = M
Part A Determine the rate law and the value of k for the following reaction using the data provided: 2NO(g) + O2(g) + 2NO2(9) [NO]; (M) [Ogl (M) Initial Rate (M4 st) 0.030 0.0055 8.55 x 103 0.030 0.0110 1.71 x 10-2 0.0055 3.42 x 10-2 0.060 Rate = 3.1 x 105 M-3 s[NO]2[0212 Rate = 57 M' s'[NO][O21 Rate = 9.4 x 103 M25 *[NO][0212 Rate = 3.8 M-1/2 s [NO](O2) 1/2 Rate = 1.7 x 103 M2 s?[NO]2[02]
The equilibrium constant, K, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) ---> CO(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.293 M COCl2, 6.15×10-2 M CO and 6.15×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 5.25×10-2mol of Cl2(g) is added to the flask? [COCl2] =___? M [CO] =___? M [Cl2] =___? M
For the following reaction, Kc = 255 at 1000 K. CO (g) + Cl2 (g) ⇌ COCl2 (g)CO (g) + Cl2 (g) ⇌ COCl2 (g) A reaction mixture initially contains a CO concentration of 0.1530 M and a Cl2Cl2 concentration of 0.176 M at 1000 K. What is the equilibrium concentration of CO at 1000 K ? What is the equilibrium concentration of Cl2 at 1000 K ? What is the equilibrium concentration of COCl2 at 1000 K ?
Using the following data, determine the rate law and calculate k: rate = k[F2]x[ClO2]y Experiment [F2] (M) [ClO2] (M) Initial rate (M/s) 1 0.40 0.05 1.2 M/s 2 0.20 0.10 0.60 M/s 3 0.40 0.10 2.4 M/s
he equilibrium constant, K, for the following reaction is 1.53×10-2 at 605 K. COCl2(g) CO(g) + Cl2(g) An equilibrium mixture of the three gases in a 8.38 L container at 605 K contains 0.288 M COCl2, 6.63×10-2 M CO and 6.63×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 14.5 L? [COCl2] = _______ M [CO] =_________ M [Cl2] =________ M