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Standard XII

Chemistry

Activation Energy

Raw milk sour...

Question

Raw milk sours in about 4 h at $27^{o} C$ , but in about 48 h in a refrigerator at $17^{o} C$ . What is the activation energy for souring of milk ?

78.3 kJ

m o l^{- 1}

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46.21 kJ

m o l^{- 1}

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23.5 kJ

m o l^{- 1}

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80.8 kJ

m o l^{- 1}

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Solution

The correct option is A 78.3 kJ $m o l^{- 1}$
$K \propto \frac{1}{t}$

$l o g \frac{K_{2}}{K_{1}} = l o g \frac{t_{1}}{t_{2}} = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} \times T_{2}})$

$l o g \frac{48}{4} = \frac{E_{a}}{8.314} (\frac{10}{300 \times 290})$

$E_{a} = 78.3 k J m o l^{- 1}$

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Raw milk sours in about 4 h at 27oC, but in about 48 h in a refrigerator at 17oC. What is the activation energy for souring of milk ?

The correct option is A 78.3 kJ mol−1K∝1tlogK2K1=logt1t2=EaR(T2−T1T1×T2)log484=Ea8.314(10300×290)Ea=78.3kJmol−1

Raw milk sours in about 4 h at $27^{o} C$ , but in about 48 h in a refrigerator at $17^{o} C$ . What is the activation energy for souring of milk ?

The correct option is A 78.3 kJ $m o l^{- 1}$
$K \propto \frac{1}{t}$

$l o g \frac{K_{2}}{K_{1}} = l o g \frac{t_{1}}{t_{2}} = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} \times T_{2}})$

$l o g \frac{48}{4} = \frac{E_{a}}{8.314} (\frac{10}{300 \times 290})$

$E_{a} = 78.3 k J m o l^{- 1}$