At 298 K , the enthalpy change ( in kJ ) for the reaction given below is CH_4(g)+O_2(g) \to C(s)+2 H_2 O(l) Given : H _2(g)+\frac{1}{2} O _2(g) \to H _2 O (l) ; ∆ H^{⊖}=-286 kJ C(s)+O_2(g) \to CO_2(g) ; ∆ H^{⊖}=-394 kJ CH_4(g)+2 O_2(g) \to CO_2(g)+2 H_2 O(l) ∆ H^{⊖}=-890 kJ

TG EAMCET 2 May 2025 Shift 1 Paper

Question : 129 of 160

Marks: +1, -0

At 298 K , the enthalpy change ( in kJ ) for the reaction given below is
CH4(g)+O2(g)→C(s)+2H2O(l)
Given : H2(g)+‌

O2(g)→H2O(l);∆H⊖=−286kJ C(s)+O2(g)‌→CO2(g);∆H⊖=−394kJ
CH4(g)+2O2(g)‌→CO2(g)+2H2O(l)∆H⊖=−890kJ

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