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Standard XII

Chemistry

Bond Order and Enthalpy

Assume each r...

Question

Assume each reaction is carried out in an open container. For which reaction will $Δ H$ = $Δ E$ ?

_{2 (g)}

+ Br

_{2 (g)}

\to

2HBr

_{(g)}

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_{(s)}

+ 2H

_{2}

_{(g)}

\to

_{2 (g)}

+ CO

_{2 (g)}

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PCl

_{5 (g)}

\to

PCl

_{3 (g)}

+ Cl

_{2 (g)}

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2CO

_{(g)}

+ O

_{2 (g)}

2CO

_{2 (g)}

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Solution

The correct option is B H $_{2 (g)}$ + Br $_{2 (g)}$ $\to$ 2HBr $_{(g)}$
We know that, $△ H = △ U + △ P V$
But for a physical or chemical change, $△ H = △ U + △ n g R T$
where, $△_{n} g$ = No. of gaseous moles of product- Reactant
So, For $△ H = △ U$ or $△ H = △ E$
$△ n g$ has to = $0$
In A option $△ n g = 2 - 2 = 0$ ; Therefore, $△ H = △ E$ .

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Similar questions

O_{(g)} + e^{-} \to O_{(g)}^{-} Δ H_{1}

O_{(g)}^{-} + e^{-} \to O_{(g)}^{- 2} Δ H_{2}

O_{2 (g)} \to O_{2 (g)}^{+} + e^{-} Δ H_{3}

O_{(g)} \to O_{(g)}^{+} + e^{-} Δ H_{4}

H_{2 (g)} \to H_{2 (g)}^{+} + e^{-} Δ H_{5}

H_{(g)} \to H_{(g)}^{+} + e^{-} Δ H_{6}

O_{(g)} + 2 e^{-} \to O^{- 2} Δ H_{7}

Which of the following option(s) is/are correct?

Q. The number of equations corresponds to the heat of formation at 298K

H_{2 (g)} + B r_{2 (l)} \to 2 H B r_{(g)} C_{(d i a m o n d)} + 2 H_{2 (g)} \to C H_{4 (g)} C_{(g r a p h i t e, s)} + 2 H_{2} (g) + \frac{1}{2} O_{2} (l) \to C H_{3} O H (l) H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)} C_{(g r a p h i t e, s)} + 3 H_{2} (g) + \frac{1}{2} O_{2} (l) \to C_{2} H_{5} O H (l) \frac{1}{2} N_{2 (g)} + \frac{3}{2} H_{2 (g)} \to N H_{3 (g)} C_{(g r a p h i t e . s)} + O_{2 (g)} \to C O_{2 (g)} C_{(d i a m o n d)} + \frac{1}{2} O_{2 (g)} \to C O (g)

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Calculate enthalpy change for the change $8 S_{(g)} \to S_{8 (g)^{'}}$ given that
$H_{2} S_{2 (g)} \to 2 H_{(g)} + 2 S_{(g)}, Δ H = 239.0 k c a l m o l^{- 1} H_{2} S_{2 (g)} \to 2 H_{(g)} + S_{(g)}, Δ H = 175.0 k c a l m o l^{- 1}$

Q. Consider the reactions

(i) C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g); K_{1} (i i) C H_{4} (g) + H_{2} O (g) ⇌ C O (g) + 3 H_{2} (g); K_{2} (i i i) C H_{4} (g) + 2 H_{2} O (g) ⇌ C O_{2} (g) + 4 H_{2} (g); K_{3}

which of the following is correct?

Q. The enthalpy of formation of reaction :

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}

What will be the enthalpy of formation of

H_{2} O (l)

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Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE ?

The correct option is B H2(g) + Br2(g) → 2HBr(g)We know that, △H=△U+△PVBut for a physical or chemical change, △H=△U+△ngRTwhere, △ng= No. of gaseous moles of product- ReactantSo, For △H=△U or △H=△E△ng has to =0In A option △ng=2−2=0 ; Therefore, △H=△E.

Assume each reaction is carried out in an open container. For which reaction will $Δ H$ = $Δ E$ ?