Milk turns sour at 40- C three times as faster as 0- C- Hence- E a in the process of turning of milk sour isA- 2-303 - 2 - 313 - 273-40log 3B- 2-303 - 2 - 313 - 273-2303 - 2 - 40-40log 1 - 3 calC- 2-303 - 2 - 40-273 - 313log 3 calD- None of these

The correct option is A 2.303× 2 × 313 × 273/40log3K_40^∘ C=Ae^ Ea/R× 313 K_0^∘ C=Ae^ Ea/R× 273 3=e^ Ea/R × (1/313 1/273) On taking logs on both sides, yo ...

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Byju's Answer

Standard VII

Chemistry

Arrhenius Equation

Milk turns so...

Question

Milk turns sour at $40^{\circ}$ C three times as faster as $0^{\circ}$ C. Hence, $E_{a}$ in the process of turning of milk sour is

\frac{2.303 \times 2 \times 313 \times 273}{40} l o g 3

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\frac{2.303 \times 2 \times 313 \times 273}{40} l o g (1 / 3) c a l

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\frac{2.303 \times 2 \times 40}{273 \times 313} l o g 3 c a l

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Solution

The correct option is A $\frac{2.303 \times 2 \times 313 \times 273}{40} l o g 3$
$K_{40^{\circ} C} = A e^{- E a / R \times 313} K_{0^{\circ} C} = A e^{- E a / R \times 273} 3 = e^{- E a / R \times (1 / 313 - 1 / 273)}$

On taking logs on both sides, you get the required answer

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Milk turns sour at 40∘C three times as faster as 0∘C. Hence, Ea in the process of turning of milk sour is

The correct option is A 2.303×2×313×27340log3K40∘C=Ae−Ea/R×313K0∘C=Ae−Ea/R×2733=e−Ea/R×(1/313−1/273) On taking logs on both sides, you get the required answer

Milk turns sour at $40^{\circ}$ C three times as faster as $0^{\circ}$ C. Hence, $E_{a}$ in the process of turning of milk sour is

The correct option is A $\frac{2.303 \times 2 \times 313 \times 273}{40} l o g 3$
$K_{40^{\circ} C} = A e^{- E a / R \times 313} K_{0^{\circ} C} = A e^{- E a / R \times 273} 3 = e^{- E a / R \times (1 / 313 - 1 / 273)}$

On taking logs on both sides, you get the required answer