Question

While studying the decompoistion of gaseous `N_(2)O_(5)`, it is observed that a plot of logarithm of its partial pressure versus time is linear. What kinetic parameters can be obtained form this observation?

Answer 1

Correct Answer - First order reaction
While studying the decompoistion of gaseous `N_(2)O_(5)`, it is observed that a plot of logarithm of its partial pressure versus time is linear. This shows that the decompoistion of `N_(2)O_(5)` follows first order kinetics.
`k = (2.303)/(t) log.(A_(0))/(A_(t))`, or `k = (2.303)/(t) log.(a)/(a-x)`
where `A_(0)` is the initial concentration of the reactants and `A_(t)` is the concentration of the reactants after time `t`. Therefore,
`log A_(t) = log A_(0) - (k)/(2.303)t` , `[[y = mx+c],["Equation of straight line"],["where m = slope"=(-k)/(2.303)],["Integrcept" (c ) = log A_(0)]]`

For the given case, the graph is between logarithm of partial pressure and time. form the slope of the graph, rate constant for the reaction can be calculated.

Answer 2

Correct Answer - First order reaction
While studying the decompoistion of gaseous `N_(2)O_(5)`, it is observed that a plot of logarithm of its partial pressure versus time is linear. This shows that the decompoistion of `N_(2)O_(5)` follows first order kinetics.
`k = (2.303)/(t) log.(A_(0))/(A_(t))`, or `k = (2.303)/(t) log.(a)/(a-x)`
where `A_(0)` is the initial concentration of the reactants and `A_(t)` is the concentration of the reactants after time `t`. Therefore,
`log A_(t) = log A_(0) - (k)/(2.303)t` , `[[y = mx+c],["Equation of straight line"],["where m = slope"=(-k)/(2.303)],["Integrcept" (c ) = log A_(0)]]`

For the given case, the graph is between logarithm of partial pressure and time. form the slope of the graph, rate constant for the reaction can be calculated.