Respuesta :
The first-order rate constant for the decomposition of N2O5
, 2N2O5(g)--> 4NO2(g)+O2(g)
at 70 C is 6.82x10^-3s-1. Suppose we start with 2.74×10−2 mol of N2O5(g) in a volume of 2.3L .
How many moles of N2O5 will remain?
The Integrate rate law can be shown as:
[tex]Kt = 2.303 log [A_0] /[A][/tex]
K is constant = [tex]6.82*10^-3 s^-1,[/tex]
t is time [tex]= 7min = 7 min * 60 ses/min = 420sec[/tex]
[A_0] is the intial mol
Molarity = [A_0] = n / V (liter) = 0.0274 /2.3 L = 0.0119 M
[tex]Kt = 2.303 log [A_0] /[A][/tex]
[tex](6.82*10^-3) * 420 = 2.303 log (0.0119) /[A][/tex]
1243 .76 = log (0.0119) – log [A][tex]1243 .76 = log (0.0119) – log [A][/tex]
[tex][A] = 0.0072 mol /L[/tex]
Thus , remaining amount of solute after 7 minute is 0.0072 mol /L
[tex]\boxed{{\text{0}}{\text{.0006789 moles}}}[/tex] of [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}[/tex] will remain after 7.0 min.
Further Explanation:
Order of reaction:
This determines dependence of rate on power of concentration of reactants involved in any chemical reaction.Reaction can be of first order, second-order, and third-order and so on.
A reaction is said to be first-order reaction if its rate varies directly with the concentration of reactants.
The first-order decomposition of [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}[/tex] occurs as follows:
[tex]{\text{2}}{{\text{N}}_{\text{2}}}{{\text{O}}_5} \to 4{\text{N}}{{\text{O}}_2} + {{\text{O}}_2}[/tex]
Since [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}[/tex] decomposes to form [tex]{\text{N}}{{\text{O}}_{\text{2}}}[/tex] and [tex]{{\text{O}}_{\text{2}}}[/tex] through first-order reaction, its rate constant can be determined with the help of below mentioned formula.
The expression to calculate rate constant for first-order reaction is as follows:
[tex]k = \dfrac{{2.303}}{t}\log \dfrac{{\left[ {{{\text{A}}_{\text{o}}}} \right]}}{{\left[ {{{\text{A}}_t}} \right]}}[/tex] …… (1)
Here,
k is rate constant.
t is the time.
[tex]\left[ {{{\text{A}}_{\text{o}}}} \right][/tex] isinitial concentration of reactant.
[tex]\left[ {{{\text{A}}_t}} \right][/tex] isconcentration of reactant at time t.
The initial concentration can be calculated as follows:
[tex]\begin{aligned}\left[ {{{\text{A}}_{\text{o}}}} \right] &= \frac{{2.74 \times {{10}^{ - 2}}{\text{ mol}}}}{{2.3{\text{ L}}}} \\&= 0.0119{\text{ M}} \\\end{aligned}[/tex]
The value of t is 7.0 min.
The value of [tex]\left[ {{{\text{A}}_{\text{o}}}} \right][/tex] is 0.0119 M.
The value of kis [tex]6.82 \times {10^{ - 3}}{\text{ }}{{\text{s}}^{ - 1}}[/tex].
Substitute 7.0 min for t, 0.0119 M for [tex]\left[ {{{\text{A}}_{\text{o}}}} \right][/tex] and [tex]6.82 \times {10^{ - 3}}{\text{ }}{{\text{s}}^{ - 1}}[/tex] for k in equation (1) to calculate [tex]\left[ {{{\text{A}}_t}} \right][/tex] for decomposition of [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}[/tex].
[tex]6.82 \times {10^{ - 3}}{\text{ }}{{\text{s}}^{ - 1}} = \dfrac{{2.303}}{{\left( {7 \times 60} \right){\text{ s}}}}\log \left( {\dfrac{{{\text{0}}{\text{.0119 M}}}}{{\left[ {{{\text{A}}_t}} \right]}}} \right)[/tex]
Solving for [tex]\left[ {{{\text{A}}_t}} \right][/tex],
[tex]\left[ {{{\text{A}}_t}} \right] = 0.0006789{\text{ M}}[/tex]
Therefore 0.0006789 moles of [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}[/tex] will remain after 7.0 minutes.
Learn more:
- Rate of chemical reaction: https://brainly.com/question/1569924
- The main purpose of conducting experiments: https://brainly.com/question/5096428
Answer Details:
Grade: Senior School
Subject: Chemistry
Chapter: Chemical Kinetics
Keywords: order, first-order, N2O5, k, t, 0.0006789 moles, NO2, O2, 7.0 minutes, NO2, O2.
