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For a decomposition of azoisopropane at ...

For a decomposition of azoisopropane at `270^@C` it was found that at t = 0 , the total pressure was found that at t = 0 , total pressure was 33.15 mm of Hg and after 3 minutes the total pressure was found to be 46.3 mm of Hg. Calculate the value of K for this reaction.
`(CH_3)_2CHN = NCH (CH_2)_2rarrN_2+C_6H_(14)`

A

`0.168 "min"^(-1)`

B

`0.173 "min"^(-1)`

C

`0.18 "min"^(-1)`

D

`0.154 "min"^(-1)`

Text Solution

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The correct Answer is:
To solve the problem of calculating the value of K for the decomposition of azoisopropane, we can follow these steps: ### Step 1: Understand the Reaction The decomposition reaction is given as: \[ (CH_3)_2CHN = NCH(CH_3)_2 \rightarrow N_2 + C_6H_{14} \] This indicates that one mole of azoisopropane decomposes to produce one mole of nitrogen gas (N2) and one mole of hexane (C6H14). ### Step 2: Initial and Final Pressures At time \( t = 0 \): - Total pressure \( P_0 = 33.15 \, \text{mm Hg} \) After 3 minutes: - Total pressure \( P_t = 46.3 \, \text{mm Hg} \) ### Step 3: Determine the Change in Pressure Let \( p \) be the pressure change due to the decomposition of azoisopropane. The pressure of the reactant at \( t = 3 \) minutes will be: \[ P_{\text{reactant}} = P_0 - p \] The total pressure at \( t = 3 \) minutes can be expressed as: \[ P_t = (P_0 - p) + p + p = P_0 + p \] This simplifies to: \[ P_t = P_0 + p \] ### Step 4: Calculate the Change in Pressure From the information provided: \[ 46.3 \, \text{mm Hg} = 33.15 \, \text{mm Hg} + p \] Solving for \( p \): \[ p = 46.3 - 33.15 = 13.15 \, \text{mm Hg} \] ### Step 5: Calculate the Pressure of the Reactant at \( t = 3 \) Minutes Now, we can find the pressure of the reactant at \( t = 3 \) minutes: \[ P_{\text{reactant}} = P_0 - p = 33.15 - 13.15 = 20 \, \text{mm Hg} \] ### Step 6: Use the First-Order Reaction Formula For a first-order reaction, the rate constant \( k \) can be calculated using the formula: \[ k = \frac{2.303}{t} \log \left( \frac{P_0}{P_t} \right) \] Where: - \( P_0 = 33.15 \, \text{mm Hg} \) - \( P_t = 20 \, \text{mm Hg} \) - \( t = 3 \, \text{minutes} \) ### Step 7: Substitute Values into the Formula Substituting the values into the formula: \[ k = \frac{2.303}{3} \log \left( \frac{33.15}{20} \right) \] ### Step 8: Calculate the Logarithm Calculating the logarithm: \[ \frac{33.15}{20} = 1.6575 \] Now, calculate \( \log(1.6575) \): \[ \log(1.6575) \approx 0.219 \] ### Step 9: Calculate \( k \) Now substituting back into the equation for \( k \): \[ k = \frac{2.303}{3} \times 0.219 \approx 0.168 \, \text{min}^{-1} \] ### Conclusion Thus, the value of \( k \) for the reaction is: \[ \boxed{0.168 \, \text{min}^{-1}} \]
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