The following reaction is first order in n2o5: n2o5(g)→no3(g)+no2(g) the rate constant for the reaction at a certain temperature is 0.053/s
calculate the rate of the reaction when [n2o5]= 5.1×10−2 m.
what would the rate of the reaction be at the same concentration as in part a if the reaction were second order?
what would the rate of the reaction be at the same concentration as in part a if the reaction were zero order?

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.

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.

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.

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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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.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years