You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Enthalpy of Neutralisation

At a particul...

Question

At a particular temperature $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l); Δ H = - 57.1 k J$ . The approximate heat evolved when $400 m L$ of $0.2$ $H_{2} S O_{4}$ is mixed with $600 m L$ of $0.1 m$ $K O H$ solution will be:

3.42kJ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

13.7 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.2kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

55kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
3.42kJ

Number of equivalent $H_{2} S O_{4}$ taken

= $\frac{0.2 \times 2 \times 400}{1000} = 0.16$

Number of equivalent $K O H$ added

= $\frac{600 \times 0.1}{1000} = 0.06$

Number of equivalent of acid and bases which neutralized each other = $0.06$

$∴$ heat evolved = $0.06 \times 57.1 k J$ = $3.426 k J$

Suggest Corrections

Similar questions

At a particular temperature $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l); Δ H = - 57.1 k J$ . The approximate heat evolved when $400 m L$ of $0.2$ $H_{2} S O_{4}$ is mixed with $600 m L$ of $0.1 m$ $K O H$ solution will be:

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^{-}

25^{o} C

is:

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. Nicotine,

C_{10} H_{14} N_{2}

, has two basic nitrogen atoms and both can react with water to give a basic solution

N i c (a q) + H_{2} O (l) ⇌ N i c H^{+} (a q) + O H^{-} (a q)

N i c H^{+} (a q) + H_{2} O (l) ⇌ N i c H_{2}^{2 +} (a q) + O H^{-} (a q)

K_{b_{1}}

8 \times 10^{- 7}

and

K_{b_{2}}

10^{- 10}

. Calculate the approximate

p H

of a

0.20

M solution.

Join BYJU'S Learning Program

At a particular temperature H+(aq) + OH−(aq) → H2O(l); ΔH = −57.1kJ. The approximate heat evolved when 400 mL of 0.2 H2SO4 is mixed with 600 mL of 0.1 m KOH solution will be:

The correct option is A 3.42kJ Number of equivalent H2SO4 taken = 0.2 × 2 × 4001000 = 0.16 Number of equivalent KOH added = 600 × 0.11000 = 0.06 Number of equivalent of acid and bases which neutralized each other = 0.06 ∴ heat evolved = 0.06 × 57.1 kJ = 3.426 kJ

At a particular temperature $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l); Δ H = - 57.1 k J$ . The approximate heat evolved when $400 m L$ of $0.2$ $H_{2} S O_{4}$ is mixed with $600 m L$ of $0.1 m$ $K O H$ solution will be: