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

Heat of Reaction

Given :C grap...

Question

Given :
$C (g r a p h i t e) + \frac{1}{2} O_{2} \to C O Δ H_{1} = - 110.5 k J / m o l$

$C O + \frac{1}{2} O_{2} \to C O_{2} Δ H = - 283.2 k J / m o l$

The heat of reaction (in $k J / m o l$ ) for the following reaction is:
$C (g r a p h i t e) + O_{2} \to C O_{2} Δ H = ?$

393.7

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

-393.7

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

-172.7

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

172.7

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

Open in App

Solution

The correct option is B -393.7
The thermochemical reactions are given as:
$C (g r a p h i t e) + \frac{1}{2} O_{2} \to C O Δ H_{1} = - 110.5 k J / m o l$

$C O + \frac{1}{2} O_{2} \to C O_{2} Δ H_{2} = - 283.2 k J / m o l$

The above two reactions are added to obtain the following reaction using Hess's law :

$C (g r a p h i t e) + O_{2} \to C O_{2} Δ H_{3} = ?$
so,
$Δ H_{3} = Δ H_{1} + Δ H_{2}$
$Δ H_{3} = - 110.5 + (- 283.2)$
$Δ H_{3} = - 393.7 k J / m o l$

Suggest Corrections

Similar questions

C (g r a p h i t e) + \frac{1}{2} O_{2} \to C O Δ H_{1} = - 110.5 k J / m o l

C O + \frac{1}{2} O_{2} \to C O_{2} Δ H = - 283.2 k J / m o l

k J / m o l

C (g r a p h i t e) + O_{2} \to C O_{2} Δ H = ?

C

+ \frac{1}{2} O_{2} \to C O; Δ H = - 110.5

C O + \frac{1}{2} O_{2} \to C O_{2}; Δ H = - 283.2

C (graphite) + O_{2} \to C O_{2}

From the thermo chemical reactions,

$C_{(g r a p h i t e)} + \frac{1}{2} O_{2} \to C O; Δ H = - 110.5 K J$

$C O + \frac{1}{2} O_{2} \to C O_{2}; Δ H = - 283.2 K J .$

$Δ H$ for the reaction, $C_{(g r a p h i t e)} + O_{2} \to C O_{2}$ is:

C + O_{2} \to C O_{2}

C + \frac{1}{2} O_{2} \to C O

C O + \frac{1}{2} O_{2} \to C O_{2}

k J / m o l

A + \frac{1}{2} O_{2} \to A O

- 50

A O + \frac{1}{2} O_{2} \to A O_{2}

100

A + O_{2} \to A O_{2}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program