CameraIcon

SearchIcon

MyQuestionIcon

1

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

Heat of Formation

Given the ent...

Question

Given the enthalpy of formation of $C O_{2}$ (g) is $- 94.0$ KJ, of CaO(s) is $- 152$ KJ, and the enthalpy of the reaction $C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$ is 42 KJ, the enthalpy of formation of $C a C O_{3} (s)$ is

A

- 268

KJ

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

B

+ 202

KJ

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

C

- 202

KJ

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

D

- 288

KJ

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

Open in App

Solution

The correct option is D $- 288$ KJ
Given:
$C a C O_{3} (s) \to C a O (s) + C O_{2} (g), Δ H = 42 K J$
$Δ H = Δ H_{f} (C O_{2}) + Δ H_{f} (C a O) - Δ H_{f} (C a C O_{3})$
$42 = - 94 + (- 152) - Δ H_{f} (C a C O_{3})$
$42 = - 246 - Δ H_{f} (C a C O_{3})$
$= Δ H_{f} (C a C O_{3})$
$- 288 K J$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The standard enthalpy of formation of octane

(C_{8} H_{18})

- 250 k J / m o l

. Calculate the enthalpy of combustion of

C_{8} H_{18}

.
Given : enthalpy of formation of

C O_{2} (g)

H_{2} O (l)

- 394 k J / m o l

- 286 k J / m o l

Paragraph for below question

नीचे दिये गये प्रश्न के लिए अनुच्छेद

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called standard molar enthalpy of formation (

Δ_{f} H^{⊖}

). Enthalpy of reaction can be calculated in terms of enthalpy of formation of the reactants and the products of the reaction.

अपने समूहन की सर्वाधिक स्थायी अवस्थाओं (संदर्भ अवस्थाएं भी कहलाती हैं) में अपने तत्वों से एक मोल यौगिक के निर्माण के लिए मानक एन्थैल्पी परिवर्तन मानक मोलर संभवन एन्थैल्पी (

Δ_{f} H^{⊖}

) कहलाती है। अभिक्रिया की एन्थैल्पी की गणना अभिक्रिया के अभिकारकों व उत्पादों की संभवन एन्थैल्पी के पदों में की जा सकती है।

Q. If heat of formation for CaCO₃(s), CaO(s) and CO₂(g) are –800 kJ mol^–1, –400 kJ mol^–1 and –250 kJ mol^–1 respectively then calculate enthalpy change of following reaction

CaCO₃(s) → CaO(s) + CO₂(g)

प्रश्न - यदि CaCO₃(s), CaO(s) तथा CO₂(g) की संभवन ऊष्मा क्रमशः –800 kJ mol^–1, –400 kJ mol^–1 तथा –250 kJ mol^–1 हैं; तो निम्नलिखित अभिक्रिया के एन्थैल्पी परिवर्तन की गणना कीजिए।

CaCO₃(s) → CaO(s) + CO₂(g)

Q. Calculate the lattice enthalpy of

C a C l_{2}

, given that :
Enthalpy of sublimation for

C a (s) \to C a (g) = 121 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} (g) \to 2 C l (g) = 242.8 k J m o l^{- 1}

Ionisation energy of

C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}

Electron gain enthalpy of

2 C l \to 2 C l^{- 1} = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}

Enthalpy of formation of

C a C l_{2} = - 795 k J m o l^{- 1}

Q. The enthalpy change for a reaction

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

is -92.2 kJ/mol. Calculate the enthalpy of formation of ammonia.

Q. Calculate the lattice enthalpy of

C a C l_{2}

, given that :
Enthalpy of sublimation for

C a (s) \to C a (g) = 121 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} (g) \to 2 C l (g) = 242.8 k J m o l^{- 1}

Ionisation energy of

C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}

Electron gain enthalpy of

2 C l \to 2 C l^{- 1} = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}

Enthalpy of formation of

C a C l_{2} = - 795 k J m o l^{- 1}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Thermochemistry

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon