Consider the reaction:
2N₂O₅ (soln) → 4NO₂ (soln) + O₂ (soln)
As the reaction is first order:
So the rate law will be :
[tex]\frac{ln([start]}{[after])}=k\times t[/tex]
[start]=0.092 mol L⁻¹
k=4.82 × 10⁻³ s⁻¹
t=151 s
Putting these values in the above equation:
[tex]\frac{ln([0.092]}{[after])}=4.82\times 10^{-3}\times 151[/tex]
[tex]ln(0,068]/[after]) = (4,28\times 10^{-3})\times 151[/tex]
[after]=0.044 M
as there is 1 L volume so number of moles remaining after 151 s is 0.044 M.
The decomposition of N₂O₅ in solution in carbon tetrachloride is first order with a rate constant of 4.82 × 10⁻³ s⁻¹. If initially there are 0.092 mol of N₂O₅ in a 1.00-L vessel, after 151 s there are 0.044 mol.
Let's consider the decomposition of dinitrogen pentoxide in solution in carbon tetrachloride.
2 N₂O₅ (soln) → 4 NO₂ (soln) + O₂ (soln)
Initially, there are 0.092 mol of N₂O₅ in a 1.00-L vessel. The initial molar concentration of N₂O₅ is:
[tex][N_2O_5]_0 = \frac{0.092mol}{1.00L} = 0.092 M[/tex]
The reaction is first order and has a rate constant (k) of 4.82 × 10⁻³ s⁻¹ at 64°C. We can calculate the concentration ([N₂O₅]) after 151 s (t) using the following expression.
[tex][N_2O_5] = [N_2O_5]_0 \times e^{-k \times t} \\\\[N_2O_5] = 0.092 M \times e^{-(4.82 \times 10^{-3}s^{-1} ) \times 151s} = 0.044 M[/tex]
The concentration of N₂O₅ after 151 s is 0.044 M. The moles of N₂O₅ in a 1.00-L vessel are:
[tex]1.00 L \times 0.044mol/L = 0.044 mol[/tex]
The decomposition of N₂O₅ in solution in carbon tetrachloride is first order with a rate constant of 4.82 × 10⁻³ s⁻¹. If initially there are 0.092 mol of N₂O₅ in a 1.00-L vessel, after 151 s there are 0.044 mol.
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