Ncert Exercises & Solutions

Q.14. Calculate the standard enthalpy of formation of $C H_{3} O H_{(f)}$ from the following data:

$C H_{3} O H_{(l)} + \frac{3}{2} O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)}; ∆_{r} H^{θ} = - 726 k J m o l^{- 1}$
$C_{(g)} + O_{2 (g)} \to C O_{2 (g)}; ∆_{c} H^{θ} = - 393 k J m o l^{- 1}$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(f)}; ∆_{f} H^{θ} = - 286 k J m o l^{- 1} .$

NEETprep Answer:

The reaction takes place during the formation of

C H_{3} O H_{(l)}

can be written as:

C_{(s)} + 2 H_{2} O_{(g)} + \frac{1}{2} O_{2 (g)} \to C H_{3} O H_{(l)} (1)

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) – equation (i)

∆_{f} H^{θ} [C H_{3} O H_{(l)}] = ∆_{c} H^{θ} + 2 ∆_{f} H^{θ} [H_{2} O_{(l)}] - ∆_{r} H^{θ}

= (- 393 k J m o l^{- 1}) + 2 (- 286 k J m o l^{- 1}) - (- 726 k J m o l^{- 1})

= (- 393 - 572 + 726) k J m o l^{- 1}

∴ ∆_{f} H^{θ} [C H_{3} O H_{(l)}] = - 239 k J m o l^{- 1}

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