Given- that- H2Ol - H- aq - OH- aq- - H - 57-32kJ H2g-12O2g - H2Ol- - H - -286-02kJThen- calculate the enthalpy of formation of OH- at 25oC is-

The correct option is A 228.8 kJ H_2(g) + 12O_2(g) → H_2O(l); Δ H = 286.02KJ Then Δ H_reaction = ∑Δ H _f(products) ∑Δ H_f(reactants) ∴Δ H_rea ...

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Standard X

Chemistry

Atom

Given, that, ...

Question

Given, that,

$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02 k J$

Then, calculate the enthalpy of formation of $O H^{-}$ at $25^{o} C$ is:

-228.8 kJ

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-343.53 kJ

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-33.53 kJ

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228.8 kJ

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Solution

The correct option is A -228.8 kJ
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02 K J$

Then $Δ H_{r e a c t i o n} = \sum Δ H_{f (p r o d u c t s)} - \sum Δ H_{f (r e a c t a n t s)}$

$∴ Δ H_{r e a c t i o n} = Δ H_{f} (H_{2} O) - Δ H_{f} (H_{2}) - \frac{1}{2} Δ H_{f} (O_{2})$

$- 286.02 = Δ H_{f} (H_{2} O) - 0 - \frac{1}{2} (0)$

$\Rightarrow Δ H_{f} (H_{2} O) = - 286.02 K J$

Now for ionization of water $(H_{2} O)$

$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 K J$

$Δ H_{r e a c t i o n} = \sum Δ H_{f (p r o d u c t s)} - \sum Δ H_{f (r e a c t a n t s)}$

$Δ H_{r e a c t i o n} = Δ H_{f} (H^{+}) + Δ H_{f} (O H^{-}) - Δ H_{f} (H_{2} O)$

$57.32 = 0 + Δ H_{f} (O H)^{-} - (- 286.02)$

$\Rightarrow - Δ H_{f} (O H^{-}) = 286.02 - 57.32$

$\Rightarrow - Δ H_{f} (O H^{-}) = 228.87$

$\Rightarrow Δ H_{f} (O H^{-}) = - 228.8 k J$

Hence, the correct option is $A$

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Similar questions

Given that

;

H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02

Then calculate the enthalpy of formation of

O H^{-}

25^{o}

Q. On the basis of the following thermochemical data:

(Δ_{f} G^{0} (H)_{(a q)}^{+} = 0)

H_{2} O (l) ⟶ H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286.20 k J

The value of enthalpy of formation of

O H^{-}

ion at

25^{o} C

is:

Q. Consider the following reactions.

(a)

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

(b)

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(l)}, Δ H = X_{2}

m o l^{- 1}

(c)

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)} - X_{3}

m o l^{- 1}

(d)

C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)} + X_{4}

m o l^{- 1}

Enthalpy of formation of

H_{2} O_{(l)}

is:

Q. Consider the following reactions :

H^{+} (a q) + O H^{-} (a q) = H_{2} O (l); Δ H = - X_{1} k J . m o l^{- 1}

II)

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H = - X_{2} k J . m o l^{- 1}

III)

C O_{2} (g) + H_{2} (g) = C O (g) + H_{2} O (l); Δ H = + X_{3} k J . m o l^{- 1}

IV)

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O (g) + H_{2} O (l); Δ H = + X_{4} k J . m o l^{- 1}

Enthalpy of formation of

H_{2} O (l)

is:

Q. Given that :

2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g)

;

Δ H = - 787

H_{2} (g) + 1_{/ 2} O_{2} (g) \to H_{2} O (l)

;

Δ

= - 286

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)

;

Δ

= - 1310

kJ.

Heat of formation of acetylene is:

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Given, that,H2O(l)→H+(aq)+OH−(aq);ΔH=57.32kJH2(g)+12O2(g)→H2O(l);ΔH=−286.02kJThen, calculate the enthalpy of formation of OH− at 25oC is:

Given, that,

$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02 k J$

Then, calculate the enthalpy of formation of $O H^{-}$ at $25^{o} C$ is: