Question

A)

The decomposition of nitramide in aqueous solution at 25°C NH NO2(aq) → N2O(g) + H2O(1) is first order in NH, NO, with a rate constant of 4.70x10s. M after 46963 s If the initial concentration of NH NO2 is 0.646 M, the concentration of NHNO, will be have passed. Submit Answer Retry Entire Group 7 more group attempts remaining

B)

C)

Answer 1

Answer:

Explanation:

A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

We can Use the following First order kinetic equation for first-order reactions:

ln [A]t / [A]0 = - kt

where k= rate, t= time, [A]t= concentration after the elapsed time and [A]0 =initial concentration.

Step 2: (B) Calculation of  the concentration at time t

Given initial concentration = [A]0 = 0.646 M K = 4.70 × 10^-5 s^-1

time(t)= 46963 s and concentration at time t = [A]t = we need to calculate

So by using the first order kinetics equation

ln [A]t / [A]0 = - kt

=> ln [A]t / [0.646 M ] = - 4.70 × 10⁵ s^-1× 46963 s

=> ln [A]t / [0.646 M ] = -2.207261

taking exponential (e) both sides of equation to remove ln

=> e ^{ln [A]t / [0.646 M ]}= e ^-2.207261

=> [A]t / [0.646 M ] = e ^-2.207261

=> [A]t = 0.1100015305 × 0.646 M = 7.11 ×10^-2 M

Step 2: (B) Calculation of  the rate constant (k)

Given initial concentration = [A]0 = 9.10 × 10^-2 M

time(t)= 809 min and concentration at time t (i.e at 809 min ) = [A]t = 2.39 × 10^-2 M

So by using the first order kinetics equation

ln [A]t / [A]0 = - kt

=> ln [ 2.39 × 10^-2 M] / [ 9.10 × 10^-2 M ] = -k × 809 min

=> -1.336981048  = -k × 809 min

=> k= 1.65× 10^-3 min^-1

Step 3: (C) Calculation of  the time (t)

Given initial concentration = [A]0 = 0.752  M K = 4.70 × 10^-5 s^-1

time(t)= we need to find and concentration at time t = [A]t = 7.90 × 10^-2 M

So by using the first order kinetics equation

ln [A]t / [A]0 = - kt

=> ln [7.90 × 10^-2 M ] / [ 0.752  M ] = - 4.70 × 10^-5 s^-1 × t

=> -2.253288471 = -4.70 × 10^-5 s^-1 × t

=> t= 47942.308 s