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The rate constant k , for the reaction N...

The rate constant k , for the reaction `N_(2)O_(5) to 2NO_(2) (g) + (1)/(2) O_(2)(g)` correspond to concentration of `N_(2)O_(5)` initially and at time t .

A

`[N_(2)O_(5)]_(t) [N_(2)O_(5)]_(0) + kt`

B

`[N_(2)O_(5)]_(0) = [N_(2)O_(5)]_(t) e^(kt)`

C

`log_(10) [N_(2)O_(5)]_(t) = log_(10)[N_(2)O_(5)]_(0) - kt`

D

ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))= kt`

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To solve the problem regarding the rate constant \( k \) for the reaction \[ N_2O_5 \rightarrow 2NO_2 + \frac{1}{2} O_2, \] we will derive the relationship between the concentration of \( N_2O_5 \) at initial time and at time \( t \). ### Step-by-Step Solution: 1. **Understanding the Reaction**: The given reaction shows the decomposition of dinitrogen pentoxide (\( N_2O_5 \)) into nitrogen dioxide (\( NO_2 \)) and oxygen (\( O_2 \)). This is a first-order reaction with respect to \( N_2O_5 \). 2. **Writing the Rate Law**: For a first-order reaction, the rate law can be expressed as: \[ -\frac{d[N_2O_5]}{dt} = k[N_2O_5]. \] 3. **Integrating the Rate Law**: To find the concentration of \( N_2O_5 \) at time \( t \), we integrate the rate equation: \[ \int \frac{1}{[N_2O_5]} d[N_2O_5] = -k \int dt. \] This gives us: \[ \ln[N_2O_5] = -kt + C, \] where \( C \) is the integration constant. 4. **Applying Initial Conditions**: At \( t = 0 \), let the initial concentration of \( N_2O_5 \) be \( [N_2O_5]_0 \). Thus, we have: \[ \ln[N_2O_5]_0 = C. \] Substituting back into our equation gives: \[ \ln[N_2O_5] = -kt + \ln[N_2O_5]_0. \] 5. **Rearranging the Equation**: We can rewrite this equation in terms of concentrations: \[ \ln\left(\frac{[N_2O_5]}{[N_2O_5]_0}\right) = -kt. \] 6. **Exponentiating Both Sides**: To eliminate the natural logarithm, we exponentiate both sides: \[ \frac{[N_2O_5]}{[N_2O_5]_0} = e^{-kt}. \] 7. **Final Expression**: Rearranging gives us the final expression for the concentration of \( N_2O_5 \) at time \( t \): \[ [N_2O_5] = [N_2O_5]_0 e^{-kt}. \] ### Conclusion: The rate constant \( k \) can be determined from the change in concentration of \( N_2O_5 \) over time using the derived formula. ---

To solve the problem regarding the rate constant \( k \) for the reaction \[ N_2O_5 \rightarrow 2NO_2 + \frac{1}{2} O_2, \] we will derive the relationship between the concentration of \( N_2O_5 \) at initial time and at time \( t \). ...
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The rate constant k , for the reaction N_(2)O_(5)(g) rarr 2NO_(2) (g) + (1)/(2) O_(2)(g) is 2.3 xx 10^(-2) s^(-1) . Which equation given below describes the change of [N_(2)O_(5)] with time ? [N_(2)O_(5)]_(0) and [N_(2)O_(5)]_(t) correspond to concentration of N_(2)O_(5) initially and at time, t ?

For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

For the first order reaction 2N_(2)O_(5)(g) rarr 4NO_(2)(g) + O_(2)(g)

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