Ncert Exercises & Solutions

The decomposition of A into product has value of k as 4.5 × 103 s^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would k be 1.5 × 10⁴s^–1?

From the Arrhenius equation, we obtain

\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

A l s o, k_{1} = 4.5 \times 10^{3} s^{- 1}

T_{1} = 273 + 10 = 283 K

k_{2} = 1.5 \times 10^{4} s^{- 1}

E_{a} = 60 K J m o l^{- 1} = 6.0 \times 10^{4} J m o l^{- 1}

T h e n,

\begin{aligned} \log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} = \frac{6.0 \times 10^{4} J {mol}^{- 1}}{2.303 \times 8.314 {JK}^{- 1} {mol}^{- 1}} (\frac{T_{2} - 283}{283 T_{2}}) \\ \Rightarrow 0.5229 = 3133.627 (\frac{T_{2} - 283}{283 T_{2}}) \\ \Rightarrow \frac{0.5229 \times 283 T_{2}}{3133.627} = T_{2} - 283 \\ \Rightarrow 0.0472 T_{2} = T_{2} - 283 \\ \Rightarrow 0.9528 T_{2} = 283 \\ \Rightarrow T_{2} = 297.019 K (approximately) \\ = 297 K \\ = 24^{\circ} C \\ Hence, k would be 1.5 \times 10^{4} s^{- 1} at 24^{\circ} C . \end{aligned}

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