Advertisements
Advertisements
Question
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
\[\ce{CH3CHO_{(g)} -> CH4_{(g)} + CO_{(g)}}\] Rate = k[CH3CHO]3/2
Advertisements
Solution
Given rate = k[CH3CHO]3/2
∴ Order of reaction = \[\ce{\frac{3}{2}}\]
Dimension of k = \[\ce{\frac{Rate}{[CH3CHO]^{3/2}}}\]
= \[\ce{\frac{(mol L^{-1} s^{-1})}{(mol L^{-1})^{3/2}}}\]
= L1/2 mol−1/2 s−1
APPEARS IN
RELATED QUESTIONS
In a first order reaction x → y, 40% of the given sample of compound remains unreacted in 45 minutes. Calculate rate constant of the reaction.
For a reaction A + B ⟶ P, the rate is given by
Rate = k [A] [B]2
What is the overall order of reaction if A is present in large excess?
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :
| t/s | 0 | 30 | 60 |
| [CH3COOCH3] / mol L–1 | 0.60 | 0.30 | 0.15 |
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
For a reaction, \[\ce{A + B -> Product}\]; the rate law is given by, r = k [A]1/2 [B]2. What is the order of the reaction?
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
\[\ce{3NO_{(g)} -> N2O_{(g)}}\] Rate = k[NO]2
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
\[\ce{H2O2_{( aq)} + 3I^-_{( aq)} + 2H^+ -> 2H2O_{(l)} + I^-_3}\] Rate = k[H2O2][I−]
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled?
Define the following term:
Pseudo first-order reaction
Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
| Experiment | Initial concentration of [A]/mol L–¹ |
Initial concentration of [B]/mol L–¹ |
Initial rate of formation of [C]/mol L–¹ s–¹ |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
The value of rate constant of a pseudo first order reaction ______.
For a complex reaction:
(i) order of overall reaction is same as molecularity of the slowest step.
(ii) order of overall reaction is less than the molecularity of the slowest step.
(iii) order of overall reaction is greater than molecularity of the slowest step.
(iv) molecularity of the slowest step is never zero or non interger.
Why is the probability of reaction with molecularity higher than three very rare?
Why can’t molecularity of any reaction be equal to zero?
Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
Assertion: Order of the reaction can be zero or fractional.
Reason: We cannot determine order from balanced chemical equation.
The rate constant for the reaction \[\ce{2H2O5 -> 4NO2 + O2}\] is 30 × 10–5 sec–1. if the rate is 204 × 10–5 mol L–1 S–1, then the concentration of N2O5 (in mol–1) is-
The following data was obtained for chemical reaction given below at 975 K.
\[\ce{2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)}\]
| [NO] | [H2] | Rate | |
| Mol L-1 | Mol L-1 | Mol L-1 s-1 | |
| (1) | 8 × 10-5 | 8 × 10-5 | 7 × 10-9 |
| (2) | 24 × 10-5 | 8 × 10-5 | 2.1 × 10-8 |
| (3) | 24 × 10-5 | 32 × 10-5 | 8.4 × 10-8 |
The order of the reaction with respect to NO is ______. (Integer answer)
A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is ______. (Use ln 2 = 0.693)
