A - B- - H - -10 kJ mol-1- Eaf - 50 kJ mol-1 then Ea of B - A will be-

The correct option is D 60 kJ mol^ 1 As given, A → B, Δ H = 10 kJ mol^ 1 it is an exothermic reaction. E_a(b) = #160;E_a(f) (Δ H) ...

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Activation Energy

A → B, Δ H = ...

Question

$A \to B, Δ H = - 10 k J m o l^{- 1}, E_{a (f)} = 50 k J m o l^{- 1} t h e n E_{a} o f B \to A w i l l b e$ :

40 k J m o l^{- 1}

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50 k J m o l^{- 1}

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- 50 k J m o l^{- 1}

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60 k J m o l^{- 1}

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Solution

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Similar questions

B + D \to E + 2 C, Δ H

Δ H (k J / m o l)

\frac{1}{2} A \to B + 50

3 B \to 3 C + D - 125

E + A \to 2 D + 350

B + D \to E + 2 C, Δ H

Consider the following reactions:

(a) $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l) Δ H = - x_{1} k J m o l^{- 1}$

(b) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - x_{2} k J m o l^{- 1}$

(d) $C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l) Δ H = x_{4} k J m o l^{- 1}$

Enthalpy of formation of $H_{2} O (l)$ is

Δ H (k J / m o l)

\frac{1}{2} A \to B

+ 150

3 B \to 2 C + D

- 125

E + A \to 2 D

+ 350

B + D \to E + 2 C, Δ H

A \to B, Δ H = 4 k c a l m o l^{- 1}, Δ S = 10 c a l m o l^{- 1} k^{- 1}

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