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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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a

-246.0 kJ mol^-1

b

40.1 kJ mor^-1

c

-85.1 kJ mol^-1

d

246.0 kJ mol^-1

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detailed solution

Correct option is C

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

The required chemical equation for which $Δ_{f} H^{\circ}$ is required is C(graphite) + 2H₂(g) ~ CH₄(g)

This equation is obtained by the manipulations Eq.(ii) + 2 Eq. (i) - Eq. (iii)

$Hence Δ_{f} H^{\circ} ({CH}_{4}, g) = [- 394.9 + 2 (- 286.1) - (- 882.0)] kJ {mol}^{- 1} = - 85.1 kJ {mol}^{- 1}$

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