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Chemistry Question on Thermodynamics terms

Standard molar enthalpies of formation of CaCO3(s)CaO(s) {CaCO_{3(s)} CaO_{(s)}} and CO2(g) {CO_{2(g)}} are −1206.92 kJ mol−1,−635.09 kJ mol−1 {-1206.92\, kJ \, mol^{-1}, -635.09\, kJ\, mol^{-1}} and −393.51 kJ mol−1 {-393.51\, kJ\, mol^{-1}} respectively. The ΔHr\Delta H_r for decomposition of CaCO3(s) {CaCO_{3(s)}} is

A

\ce178.3 kJ mol−1\ce{178.3\, kJ\, mol^{-1}}

B

\ce−178.3 kJ mol−1\ce{ - 178.3\, kJ\, mol^{-1}}

C

\ce1448.5 kJ mol−1\ce{1448.5\, kJ\, mol^{-1}}

D

\ce−1448.5 kJ mol−1\ce{ - 1448.5\, kJ\, mol^{-1}}

Answer

\ce178.3 kJ mol−1\ce{178.3\, kJ\, mol^{-1}}

Explanation

Solution

CaCO3(s)−>CaO(s)+CO2(g) {CaCO_{3(s)} -> CaO_{(s)} + CO_{2(g)}}
ΔrH∘=ΔfH∘(CaO)+ΔfH∘(CO2)−ΔfH∘(CaCO3)\Delta_r H^\circ = \Delta_f H^\circ (CaO) + \Delta_f H^\circ (CO_2) - \Delta_f H^\circ (CaCO_3)
? ? ? ? ? ? = (- 635.09) + (- 393.51)-(-,1206.92)
? ? ? ? ? ? = 178.3 kJ mol−1 {178.3\, kJ\, mol^{-1}}