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Standard XII

Chemistry

Relation between Kp and Kc

The molar ent...

Question

The molar enthalpies of combusion of $C_{2} H_{2 (g)}$ , $C_{(g r a p h i t e)}$ and $H_{2 (g)}$ are -1300,-394,-286 kJmol $^{- 1}$ respectively. The standard enthalpy of formation of $C_{2} H_{2 (g)}$ is

-226 kJmol

^{- 1}

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-626 kJmol

^{- 1}

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226 kJmol

^{- 1}

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626 kJmol

^{- 1}

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Solution

The correct option is D 626 kJmol $^{- 1}$
(i) C (graphite) $+ O_{2} \to C O_{2} - 394$
(ii) $H_{2} + \frac{1}{2} O_{2} \to H_{2} O - 286$
(iii) $C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2} O - 1300$
2 x (i) + (ii) + (iii)
2 C (graphite) $+ H_{2} (g) \to C_{2} H_{2}$
$Δ H = - 2 \times 394 - 286 + 1300$
$Δ H = 226 K J$ mol $^{- 1}$

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The molar enthalpies of combusion of C2H2(g),C(graphite) and H2(g)are -1300,-394,-286 kJmol−1 respectively. The standard enthalpy of formation of C2H2(g) is

The correct option is D 626 kJmol−1(i) C (graphite) +O2→CO2−394(ii) H2+12O2→H2O−286(iii) C2H2+52O2→2CO2+H2O−13002 x (i) + (ii) + (iii)2 C (graphite) +H2(g)→C2H2ΔH=−2×394−286+1300ΔH=226KJ mol−1

The molar enthalpies of combusion of $C_{2} H_{2 (g)}$ , $C_{(g r a p h i t e)}$ and $H_{2 (g)}$ are -1300,-394,-286 kJmol $^{- 1}$ respectively. The standard enthalpy of formation of $C_{2} H_{2 (g)}$ is