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The first-order rate constant for the decomposition of n2o5, 2n2o5(g)→4no2(g)+o2(g) at 70∘c is 6.82×10−3 s−1. suppose we start with 2.00×10−2 mol of n2o5(g) in a volume of 2.3 l . you may want to reference (page) section 14.4 while completing this problem. part a how many moles of n2o5 will remain after 4.0 min ?,

Respuesta :

The rate constant for 1st order reaction is

K = (2.303 /t) log (A0 /A)

Where, k is rate constant

t is time in sec

A0 is initial concentration

(6.82 * 10-3) * 240 = log (0.02 /A)

1.63 = log (0.02 /A)

-1.69 – log A = 1.63

Log A = - 0.069

A = 0.82

Hence, 0.82 mol of A remain after 4 minutes.

ajeigbeibraheem

The moles of N2O5 that will remain after 4.0 mins is 0.00389 moles

From the given information, the rate law for a first order reaction can be computed as:

[tex]\mathbf{K = \dfrac{1}{t} In (\dfrac{A_o}{A_t}) }[/tex]

where;

  • rate constant (K) = 6.82 × 10⁻³ s⁻¹, and;
  • time (t) = 4.0 min
  • initial number of moles A₀ = 2.00 × 10⁻²
  • number of moles at time (t) [tex]A_t[/tex] = ???

∴

the number of moles of N₂O₅ at time (t) is:

[tex]\mathbf{6.82 \times 10^{-3 }= \dfrac{1}{4 \ min \times \dfrac{60\ s}{1 \ min}} \times In (\dfrac{2.00 \times 10^{-2}}{A_t})}[/tex]

[tex]\mathbf{1.6368 = In (\dfrac{2.00 \times 10^{-2}}{A_t})}[/tex]

[tex]\mathbf{ (A_t)= 2.00 \times 10^{-2} \times e^{-(1.6368)}}[/tex]

[tex]\mathbf{ (A_t)=0.00389 \ moles}[/tex]

Learn more about first order rate constant here:

https://brainly.com/question/14056054?referrer=searchResults

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