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

Calorimetry

The heat of c...

Question

The heat of combustion of ethanal determined in a bomb calorimeter is $- 670.48 K$ Cals ${m o l e}^{- 1}$ at $25^{o} C$ .What is $Δ H$ at $25^{o} C$ for the reaction:

- 335.24 K C a l s .

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

- 671.08 K C a l s .

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

- 670.48 K C a l s .

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

+ 670.48 K C a l s .

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

Open in App

Solution

The correct option is C $- 670.48 K C a l s .$
Given $Δ H_{c o m b u s t i o n} (C H_{3} C H O) = - 670.48$ $K c a l s m o l^{-}$
$C H_{3} C H O + 3 O_{2} ⟶ 2 C O_{2} + 2 H_{2} O$

$Δ H_{r e a c t i o n} = \sum Δ H_{c o m b u s t i o n}^{r e a c t a n t} - \sum Δ H_{c o m b u s t i o n}^{p r o d u c t}$

$Δ H_{r e a c t i o n} = Δ H_{c o m b u s t i o n}^{C H_{3} C H O} + 3 \times 0 - 2 \times 0 - 2 \times 0$

$= - 670.48$ $K c a l s m o l^{-}$

Suggest Corrections

Similar questions

- 670.48 k c a l

{m o l}^{- 1}

298 K

Δ U

298 K

3263.9 k J m o l^{- 1}

25^{o} C

1.89 g

25^{o} C

18.94 k g

0.632^{o} C

25^{o} C

0.998 c a l / g / d e g

C_{2} H_{5} O H (l) + 3 O_{2} (g) ⟶ 2 {C O}_{2} (g) + 3 H_{2} O (l)

1364.47 k J {m o l}^{- 1}

25^{o} C

Δ_{c} H

(R = 8.314 k J {m o l}^{- 1})

1.89 g

25^{o} C

18.94 k g

{0.632}^{o} C

25^{o} C

0.998 c a l g^{- 1}

{d e g}^{- 1}

Join BYJU'S Learning Program