Answer
Verified
376.5k+ views
Hint: A first order reaction is the reaction where the concentration of the reacting species is raised to the power of one. The rate constant of any reaction is the constant of proportionality that is introduced in the rate law, that rate of any reaction is directly proportional to the concentration of reactants raised to the stoichiometric coefficients.
Complete answer:
We have been given the decomposition of a gas ${{N}_{2}}{{O}_{5}}$ that follows the first order reaction. The concentration is given in units of pressure, where after 30 minutes the pressure becomes 284.5 mm of Hg and on complete decomposition the total pressure is 584.5 mm of Hg. We have to find the rate constant.
So, first giving the concentration, variables according to the reaction as:
$\begin{align}
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g) \\
& initial\,pressure\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
& at\,time\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\
& At\,complete\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a \\
& dissociation \\
\end{align}$
So, total number of moles at time, t = $a-x+2x+\dfrac{x}{2}=a+\dfrac{3}{2}x$
After complete dissociation the total moles will be = $2a+\dfrac{a}{2}=\dfrac{5}{2}a$
As, we know the number of moles are directly proportional to the pressure, so,
$\dfrac{5}{2}a\propto 584.5mm\,Hg....(1)$
$a+\dfrac{3}{2}x\propto 284.5mm\,Hg.......(2)$
After solving equations 1 and 2 we have, a = 233.8 mm Hg and x = 33.8 mm Hg. Substituting the values of a and x (concentration at time t and final concentration after dissociation) in the first order equation we have,
$K=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$
$K=\dfrac{2.303}{30\,\min }\log \dfrac{233.8}{233.8-33.8}$
K = $5.21\times {{10}^{-3}}{{\min }^{-1}}$
So, the rate constant of the given reaction is $5.21\times {{10}^{-3}}{{\min }^{-1}}$.
Note:
The integrated form of the rate equation is used here that has rate constant equal to the negative log by time into the final concentration after decomposition upon initial concentration at time t. For graphical representation of concentration versus time, the negative slope of the graph gives us the rate constant.
Complete answer:
We have been given the decomposition of a gas ${{N}_{2}}{{O}_{5}}$ that follows the first order reaction. The concentration is given in units of pressure, where after 30 minutes the pressure becomes 284.5 mm of Hg and on complete decomposition the total pressure is 584.5 mm of Hg. We have to find the rate constant.
So, first giving the concentration, variables according to the reaction as:
$\begin{align}
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g) \\
& initial\,pressure\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
& at\,time\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\
& At\,complete\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a \\
& dissociation \\
\end{align}$
So, total number of moles at time, t = $a-x+2x+\dfrac{x}{2}=a+\dfrac{3}{2}x$
After complete dissociation the total moles will be = $2a+\dfrac{a}{2}=\dfrac{5}{2}a$
As, we know the number of moles are directly proportional to the pressure, so,
$\dfrac{5}{2}a\propto 584.5mm\,Hg....(1)$
$a+\dfrac{3}{2}x\propto 284.5mm\,Hg.......(2)$
After solving equations 1 and 2 we have, a = 233.8 mm Hg and x = 33.8 mm Hg. Substituting the values of a and x (concentration at time t and final concentration after dissociation) in the first order equation we have,
$K=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$
$K=\dfrac{2.303}{30\,\min }\log \dfrac{233.8}{233.8-33.8}$
K = $5.21\times {{10}^{-3}}{{\min }^{-1}}$
So, the rate constant of the given reaction is $5.21\times {{10}^{-3}}{{\min }^{-1}}$.
Note:
The integrated form of the rate equation is used here that has rate constant equal to the negative log by time into the final concentration after decomposition upon initial concentration at time t. For graphical representation of concentration versus time, the negative slope of the graph gives us the rate constant.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE