The decomposition of A into product has ...

NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

Question : 28 of 39

Marks: +1, -0

The decomposition of A into product has value of k as 4.5 × 10

^{3}

^{-1}

at 10°C and energy of activation 60 kJ mol. At what temperature would k be 1.5 × 10

^{4}

^{-1}

Solution:

Given

k_{1}=4.5 \times 10^{3} \mathrm{s}^{-1}

T_{1}=10+273 \mathrm{K}

\;\;\;=283 \; \mathrm{K} ;

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

T_{2}=?, E_{a}=60 \; \mathrm{kJ} \; \mathrm{mol}^{-1}

Applying Arrhenius equation,

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

\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}

= \frac{60000}{2.303 \times 8.314} \left( \frac{T_{2}-283}{283 T_{2}} \right)

or,

\log 3.333 = 3133.63 \left( \frac{T_{2}-283}{283 T_{2}} \right)

or,

\frac{0.5228}{3133.63} = \frac{T_{2}-283}{283 T_{2}}

or,

0.0472 T_{2} = T_{2} - 283

0.9528 T_{2} = 283

or,

T_{2} = \frac{283}{0.9528} = 297 \; \mathrm{K}

=297-273=24^{\circ} \mathrm{C}

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