Answer:
1) 2.703 x 10⁻³ M/s.
2) 1.378 x 10⁻⁴ M/s.
3) 0.053 M/s.
Explanation:
1) Calculate the rate of the reaction when [N₂O₅] = 5.1×10⁻² M.
The rate of the reaction = k[N₂O₅],
where, k is the rate constant of the reaction (k = 0.053 s⁻¹).
[N₂O₅] is the concentration of the reactant ([N₂O₅] = 5.1 x 10⁻² M).
∴ The rate of the reaction = k[N₂O₅] = (0.053 s⁻¹)(5.1 x 10⁻² M) = 2.703 x 10⁻³ M/s.
2) What would the rate of the reaction be at the same concentration as in Part A if the reaction were second order?
The rate of the reaction = k[N₂O₅]²,
where, k is the rate constant of the reaction (k = 0.053 M⁻¹s⁻¹).
[N₂O₅] is the concentration of the reactant ([N₂O₅] = 5.1 x 10⁻² M).
∴ The rate of the reaction = k[N₂O₅]² = (0.053 M⁻¹s⁻¹)(5.1 x 10⁻² M)² = 1.378 x 10⁻⁴ M/s.
3) What would the rate of the reaction be at the same concentration as in Part A if the reaction were zero order?
The rate of the reaction = k[N₂O₅]⁰,
where, k is the rate constant of the reaction (k = 0.053 M/s).
[N₂O₅] is the concentration of the reactant ([N₂O₅] = 5.1 x 10⁻² M).
∴ The rate of the reaction = k[N₂O₅]⁰ = (0.053 M/s)(5.1 x 10⁻² M)⁰ = 0.053 M/s.
When [N₂O₅] = 5.1 × 10⁻² M, the first-order decomposition of N₂O₅ has a reaction rate of 2.7 × 10⁻² M/s. If it were of second order, the reaction rate would be 1.4 × 10⁻⁴ M/s. If it were of zero order, the reaction rate would be 0.053 M/s.
Let's consider the following reaction.
N₂O₅(g) → NO₃(g) + NO₂(g)
This reaction is of first order with a rate constant (k) of 0.053/s. We can calculate the rate of the reaction when [N₂O₅] = 5.1 × 10⁻² M using the first-order rate law.
[tex]rate = k \times [N_2O_5] = 0.053 s^{-1} \times (5.1 \times 10^{-2}M ) = 2.7 \times 10^{-3} M/s[/tex]
For the same concentration and numerical value of the rate constant, if the reaction were of second order, we would use the second-order rate law.
[tex]rate = k \times [N_2O_5]^{2} = 0.053 M^{-1} s^{-1} \times (5.1 \times 10^{-2}M )^{2} = 1.4 \times 10^{-4} M/s[/tex]
For the same concentration and numerical value of the rate constant, if the reaction were of zero order, we would use the zero-order rate law.
[tex]rate = k \times [N_2O_5]^{0} = 0.053 M s^{-1} \times (5.1 \times 10^{-2}M )^{0} = 0.053 M/s[/tex]
When [N₂O₅] = 5.1 × 10⁻² M, the first-order decomposition of N₂O₅ has a reaction rate of 2.7 × 10⁻² M/s. If it were of second order, the reaction rate would be 1.4 × 10⁻⁴ M/s. If it were of zero order, the reaction rate would be 0.053 M/s.
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