For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t (sec) | P(mm of Hg) |

0 | 35.0 |

360 | 54.0 |

720 | 63.0 |

Calculate the rate constant.

When t = 0, the total partial pressure is P_{0} = 35.0 mm of Hg

When time t = t, the total partial pressure is P_{t} = P_{0} + p

P_{0}-p = P_{t}-2p, but by the above equation, we know p = P_{t}-P_{0}

Hence, P_{0}-p = P_{t}-2( P_{t}-P_{0})

Thus, P_{0}-p = 2P_{0} – P_{t}

We know that time

Where, k- rate constant

[R]_{°} -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

→ equation 1

At time t = 360 s, P_{t} = 54 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒

⇒ k = 2.175 × 10^{-3} s^{-1}

At time t = 720 s, P_{t} = 63 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒

Thus, k = 2.235 × 10^{-3} s^{–1}

Taking average, k = (2.235 × 10^{-3} s^{–1} + 2.175 × 10^{-3} s^{–1})/2

∴ k = 2.21 × 10^{-3} s^{–1}.

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