For the reaction
NO2(g) + CO(g) → NO(g) + CO2(g)
calculate the order of the reaction with respect to the following reactants according to the following experimental data:
Part 1 (1 point)
Order of the reaction with respect to NO2:
Part 2 (1 point)
Order of the reaction with respect to CO:
Firstly we will experiment 2 and 3
While keeping concentration of constant
The concentration of no2 is twiced and the rate of reaction becomes 4 times so we could conclude the rate is proportional to [no2]² hence
reaction order wrt no2 is 2
Now we observe experiment 1 and 2
Concentration of no2 is constant while co is changed but rate remains constant this means that rate of reaction doesn't depends upon co concentration
reaction order wrt co is 0
Reaction Equation: NO2 (g) + CO(g) ⟶ NO(g) + CO2 (g) Experiment Number Inital concentration of [NO2] (M) Inital concentration of [CO] (M) Initial rate (M/s) 1 0.200 0.200 5.00X10-4 2 0.200 0.800 8.00X10-3 3 0.600 0.200 5.00X10-4 A.) Determine the order (0, 1, or 2) with respect to NO2. B.) Determine the order (0, 1, or 2) with respect to CO.
The rate of the reaction: NO2(g) + CO(g) → NO(g) + CO2(g) was determined in three experiments at 225°C. The results are given in the following table: Experiment NO2(M) CO (M) Initial Rate –ΔNO2/Δt (M/s) 1 0.277 0.898 0.19 2 0.277 0.449 0.19 3 0.462 0.449 0.576 Calculate the value of the rate constant at 225°C using reaction 1 data
The reaction rate of CO and NO2 in the reaction CO(g) + NO2(g) → CO2(g) + NO(g) is measured using the initial rates method. The results are tabulated below. [CO] (mol/L) NO2 (mol/L) -([CO]/Δt (mol/L·s) 8.00 10-4 5.50 10-4 8.40 10-8 8.00 10-4 7.78 10-4 1.68 10-7 1.60 10-3 5.50 10-4 1.68 10-7 Determine the rate expression and calculate the rate constant for the reaction.
If the mechanism behind the reaction NO2(g) + CO(g) Ó NO(g) + CO2(g) is : 1- 2NO2(g) à 2NO(g) + O2(g) (slow) 2- NO(g) + CO(g) + O2(g) à NO2(g) + CO2(g) (fast) Then its rate law is: A) Rate = k [NO2] . [CO] B) Rate = k [NO2] . [CO2] C) Rate = k [NO 212 D) Rate = k [co]2
The rate of the reaction: CO (g) + NO2 (g) à CO2 (g) + NO (g) was measured at several temperatures, and the following data were collected: Temp (oC) K (M-1s-1) 35 0.184 45 0.322 Using this data determine the value of Ea (energy of activation)
Is no2(g)+co(g)→no(g)+co2(g) an elementary reaction? Rate = k[NO2]^2 Please explain if possible why It is or is not an elementary reaction
Be sure to answer all parts. The reaction between NO2 and CO to produce NO and CO2 is thought to occur in two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 The experimental rate law is rate = k[NO2]2. (a) Write the equation for the overall reaction. Do not include phase abbreviations. (b) Identify the intermediate(s). NO2 NO3 CO2 CO NO (c) Identify the rate-determining step. (Step 1, Step 2, Neither...
You have the following reaction: NO2 (g) + CO (g) ⟶ NO (g) + CO2 (g) The rate constant (k) at 701 K is 2.57 M-1s-1. If the activation energy is 150 kJ/mol, what is k at 895 K? R = 8.314 J/(mol*K) A) 680 M-1s-1 B) 443 M-1s-1 C) 2.58 M-1s-1 D) 0.950 M-1s-1 E) 6.52 M-1s-1
Experimental data is collected for the reaction shown below, with the following rate law: rate=k[NO2]2. What are the units of the rate constant for the reaction? NO2(g)+CO(g)→NO(g)+CO2(g) Trial123[NO2] (mol/L)0.060.060.09[CO] (mol/L)0.060.090.06Rate(mol L−1s−1)1.5408×10−61.5408×10−63.4668×10−6
The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 200 kJ/mol and the change in enthalpy for the reaction is ΔH = -200 kJ/mol . What is the activation energy for the reverse reaction?