The data are obtained for the initial rate of the reaction:
2NO(g) + O2(g) = 2NO2(g)
Data:
Run 1: [NO], M = 0.20 and [O2], M = 0.20 and Initial rate, mol/L*s = 4.64 X 10^-8
Run 2: [NO], M = 0.20 and [O2], M = 0.40 and Initial rate, mol/L*s = 9.28 X 10^-8
Run 3: [NO], M = 0.40 and [O2], M = 0.40 and Initial rate, mol/L*s = 3.71 X 10^-7
Run 4: [NO], M = 0.10 and [O2], M = 0.10 and Initial rate, mol/L*s = 5.80 X 10^-9
A. What is the rate law equation for the reaction?
B. What is the value of the specific rate law constant, including units?
C. This mechanism is suggested:
1. NO (g) + O2 (g) = NO3 (g) (equilibrium)
2. NO3 (g) + NO (g) = 2NO2 (g) (slow)
Show how this mechanism leads to the observed rate equation.
D. The reaction rate is increased five-fold when the temperature is increased from 1400 K to 1500 K. What is the activation energy for the reaction?
I really need help with this question on my study guide. Thanks in advance!!
2NO(g) + O2(g) = 2NO2(g)
Data:
Run 1: [NO], M = 0.20 and [O2], M = 0.20 and Initial rate, mol/L*s = 4.64 X 10^-8
Run 2: [NO], M = 0.20 and [O2], M = 0.40 and Initial rate, mol/L*s = 9.28 X 10^-8
Run 3: [NO], M = 0.40 and [O2], M = 0.40 and Initial rate, mol/L*s = 3.71 X 10^-7
Run 4: [NO], M = 0.10 and [O2], M = 0.10 and Initial rate, mol/L*s = 5.80 X 10^-9
A. What is the rate law equation for the reaction?
B. What is the value of the specific rate law constant, including units?
C. This mechanism is suggested:
1. NO (g) + O2 (g) = NO3 (g) (equilibrium)
2. NO3 (g) + NO (g) = 2NO2 (g) (slow)
Show how this mechanism leads to the observed rate equation.
D. The reaction rate is increased five-fold when the temperature is increased from 1400 K to 1500 K. What is the activation energy for the reaction?
I really need help with this question on my study guide. Thanks in advance!!
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A- We use the first 3 runs to find the order of reaction:
When [O2] is doubled from 1 to 2, reaction rate doubled, so O2 is first order wrt (with respect to) the reaction
When [NO) was doubled from 2 to 3, the reaction rate quadrupled, so NO is second order wrt to the reaction
rate Law: =k[NO]^2[O2]
B- pick one of the runs: you have the rate, and you have the concentrations. substitute the values to solve for K. the unit is going to be M^-2 s^-1.
C- the rate law always depends on the slow step (rate determining step). its number 2, and since its an elementary step, the molar ratio is equal to the order wrt to the reaction. i.e. NO is first order and NO3 is first order. NO3 is an intermediate and so it cannot be shown in the rate equation. we substitute it for the reaction from part A. NO3 = first order NO + first order O2. so we have 2 first order NO and 1 first order O2. so they add up to what we have got in A.
D. Use the arrhenius equation (rearranged): Ln(k2/k1)= (Ea/R) x (1/T1 - 1/T2) where Ea is activation energy, k2 and k1 are the reaction rates, R is gas consant 8.31 j/mol k and T is the temp in kelvin. Since the reaction rate went fivefolds up, use ln(5) as ln (k2/k1) and substitute and work from there.
When [O2] is doubled from 1 to 2, reaction rate doubled, so O2 is first order wrt (with respect to) the reaction
When [NO) was doubled from 2 to 3, the reaction rate quadrupled, so NO is second order wrt to the reaction
rate Law: =k[NO]^2[O2]
B- pick one of the runs: you have the rate, and you have the concentrations. substitute the values to solve for K. the unit is going to be M^-2 s^-1.
C- the rate law always depends on the slow step (rate determining step). its number 2, and since its an elementary step, the molar ratio is equal to the order wrt to the reaction. i.e. NO is first order and NO3 is first order. NO3 is an intermediate and so it cannot be shown in the rate equation. we substitute it for the reaction from part A. NO3 = first order NO + first order O2. so we have 2 first order NO and 1 first order O2. so they add up to what we have got in A.
D. Use the arrhenius equation (rearranged): Ln(k2/k1)= (Ea/R) x (1/T1 - 1/T2) where Ea is activation energy, k2 and k1 are the reaction rates, R is gas consant 8.31 j/mol k and T is the temp in kelvin. Since the reaction rate went fivefolds up, use ln(5) as ln (k2/k1) and substitute and work from there.