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Question

Given the enthalpy of formation of CO2(g) is 94.0 KJ, of CaO(s) is 152 KJ, and the enthalpy of the reaction CaCO3(s)CaO(s)+CO2(g) is 42 KJ, the enthalpy of formation of CaCO3(s) is

A
268 KJ
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B
+202 KJ
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C
202 KJ
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D
288 KJ
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Solution

The correct option is D 288 KJ
Given:
CaCO3(s)CaO(s)+CO2(g),ΔH=42KJ
ΔH=ΔHf(CO2)+ΔHf(CaO)ΔHf(CaCO3)
42=94+(152)ΔHf(CaCO3)
42=246ΔHf(CaCO3)
=ΔHf(CaCO3)
288KJ

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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 (ΔfH). 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 (ΔfH) 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)


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