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Standard XII

Chemistry

Hess' Law

The standard ...

Question

The standard enthalpy of formation of water liquid is 285.76 kJ at 298 K. Calculate the value at 373K. The molar heat capacities at constant pressure $(C_{P})$ in the given temperature range of $H_{2} (g), O_{2} (g)$ and $H_{2} O (l)$ are respectively 38.83, 29.16 and 75.312 $JK^{1}mol^{1}

Δ H_{373}^{o} (H_{2} O, (l)) = - 284.11 k J

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Δ H_{373}^{o} (H_{2} O, (l)) = + 284.11 k J

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Δ H_{373}^{o} (H_{2} O, (l)) = 28.411 k J

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None of these

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Solution

The correct option is A $Δ H_{373}^{o} (H_{2} O, (l)) = - 284.11 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ C_{P_{r e a c t i o n}} = C_{P_{H_{2} O (l)}} - C_{P_{H_{2} (g)}} - \frac{1}{2} C_{P_{O_{2} (g)}}$
$= 75.312 - 38.83 - \frac{1}{2} \times 29.16$
$Δ C_{P_{r e a c t i o n}} = 21.90 k J$
$Δ H_{373} = Δ H_{298} + n C_{P} Δ T$
$= (- 285.76 + 1 \times 21.9 \times 75 \times 10^{- 3}) k J$
$Δ H_{373} = - 284.12 k J$

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