Given are the following data at 298 KΔcH∘H2,g=−286.1kJmol−1; ΔcH∘(C, graphite )=−394.9kJmol−1 ; ΔcH∘CH4,g=−882.0kJmol−1The value of Δ4H∘CH4,g will be

(i) H2(g)+12O2(g)⟶H2O(l); ΔcH∘=−286.1kJmol−1 (ii) C (graphite) +O2 (g) ⟶CO2(g); ΔcH∘=−394.9kJmol−1 (iii) CH4(g)+2O2⟶CO2(g)+2H2O(l); ΔcH∘=−882.0kJmol−1The required chemical equation for which ΔfH∘ is required is C(graphite) + 2H2 (g) ~ CH4(g)This equation is obtained by the manipulations Eq.(ii) + 2 Eq. (i) - Eq. (iii)Hence ΔfH∘CH4,g=[−394.9+2(−286.1)−(−882.0)]kJmol−1=−85.1kJmol−1

Given are the following data at 298 KΔcH∘H2,g=−286.1kJmol−1; ΔcH∘(C, graphite )=−394.9kJmol−1 ; ΔcH∘CH4,g=−882.0kJmol−1The value of Δ4H∘CH4,g will be

(i) H2(g)+12O2(g)⟶H2O(l); ΔcH∘=−286.1kJmol−1 (ii) C (graphite) +O2 (g) ⟶CO2(g); ΔcH∘=−394.9kJmol−1 (iii) CH4(g)+2O2⟶CO2(g)+2H2O(l); ΔcH∘=−882.0kJmol−1The required chemical equation for which ΔfH∘ is required is C(graphite) + 2H2 (g) ~ CH4(g)This equation is obtained by the manipulations Eq.(ii) + 2 Eq. (i) - Eq. (iii)Hence ΔfH∘CH4,g=[−394.9+2(−286.1)−(−882.0)]kJmol−1=−85.1kJmol−1

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Given are the following data at 298 K
$Δ_{c} H^{\circ} (H_{2}, g) = - 286.1 kJ {mol}^{- 1};$ $Δ_{c} H^{\circ} (C, graphite) = - 394.9 kJ {mol}^{- 1}$ ; $Δ_{c} H^{\circ} ({CH}_{4}, g) = - 882.0 kJ {mol}^{- 1}$
The value of $Δ_{4} H^{\circ} ({CH}_{4}, g)$ will be

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-246.0 kJ mol^-1

40.1 kJ mor^-1

-85.1 kJ mol^-1

246.0 kJ mol^-1

answer is C.

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Detailed Solution

$(i) H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ_{c} H^{\circ} = - 286.1 kJ {mol}^{- 1} (ii) C (graphite) + O_{2} (g) ⟶ {CO}_{2} (g); Δ_{c} H^{\circ} = - 394.9 kJ {mol}^{- 1} (iii) {CH}_{4} (g) + 2 O_{2} ⟶ {CO}_{2} (g) + 2 H_{2} O (l); {ΔcH}^{\circ} = - 882.0 kJ {mol}^{- 1}$