Why is log Kp 2 - Kp 1 negative when Kp 2- Kp 1 -

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Properties of Determinants

Why is log Kp...

Question

Why is log(Kp₂/Kp₁) negative when Kp2<Kp1?

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Solution

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K_{c}

K_{p} .

K_{p}

$A (s) ⇌ B (g) + C (g); K_{P 1} = x a t m^{2}$

$D (s) ⇌ C (g) + E (g); K_{P 2} = x a t m^{2}$

X (s) ⇌ A (g) + 2 B (g)

Y (s) ⇌ 2 B (g) + C (g)

K_{P_{1}} = 9 \times 10^{- 3} a t m^{3}

K_{P_{1}}

K_{P_{2}} = 2

X

Y

A (s) ⇌ B (g) + C (g); K_{p_{1}} = x a t m^{2}

D (s) ⇌ C (g) + E (g); K_{p_{2}} = y a t m^{2}

α

α^{^{'}}

P^{^{'}}

A (s) ⇌ 2 B (g) + C (g)

K_{P_{1}} = 8 \times 10^{- 2}

A^{^{'}} (s) ⇌ 2 B (g) + D (g)

K_{P_{2}} = 2 \times 10^{- 2}

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