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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1
The enthalpy changes for two reactions are given by the equations. 2Cr(g)+32O2(g)→Cr2O3(g),  ΔH=−1130kJ      C(g)+12O2(g)→CO(g),  ΔH=−110kJ    2C{r_{\left( g \right)}} + \frac{3}{2}{O_{2\left( g \right)}} \to C{r_2}{O_{3\left( g \right)}},\,\,\Delta H = - 1130kJ\,\,\,\,\,\,{C_{\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \to C{O_{\left( g \right)}},\,\,\Delta H = - 110kJ\,\,\,\,
What is the enthalpy change (in kJ) for the reaction? 3C(s)+Cr2O3(s)→2Cr(s)+3CO(g)3C\left( s \right) + C{r_2}{O_3}\left( s \right) \to 2Cr\left( s \right) + 3CO\left( g \right)
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Correct option: B
+ 800kJ
Option B (+ 800kJ) is correct because the enthalpy change for the reaction is calculated by manipulating the given equations using Hess's Law.

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