The molar heat of formation of NH 4 NO 3 s is -367-5 KJ and those of N 2 O g and H 2 O l are -81-46 KJ and -285-78 KJ respectively at25- C and 1 atmospheric pressure- The U for the reaction NH 4 NO 3 s - N 2 O g -2 H 2 O l A- -125-03 kJB- 125-03 kJC- 120 kJD- -120 kJ

The correct option is A 125.03 kJ The change in enthalpy can be given by: H^o = H^o_f(products) H^o_f(reactants) =[ H^o_f(N_2O)+2 × H^o_f(H_2O)] [ H^o_f(NH_ ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity at Constant Pressure

The molar hea...

Question

The molar heat of formation of $N H_{4} N O_{3} (s)$ is $- 367.5 K J$ and those of $N_{2} O (g)$ and $H_{2} O (l)$ are $+ 81.46 K J$ and $- 285.78 K J$ respectively at $25^{o} C$ and 1 atmospheric pressure. The $△$ U for the reaction
$N H_{4} N O_{3} (s) \to N_{2} O (g) + 2 H_{2} O (l)$

-125.03 kJ

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120 kJ

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-120 kJ

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Solution

The correct option is A
-125.03 kJ

The change in enthalpy can be given by:
$△ H^{o} = △ H_{f (p r o d u c t s)}^{o} - △ H_{f (r e a c t a n t s)}^{o}$
$= [△ H_{f (N_{2} O)}^{o} + 2 \times △ H_{f (H_{2} O)}^{o}] - [△ H_{f (N H_{4} N O_{3})}^{o}]$
$= 81.46 + 2 \times (- 285.78) - (- 367.5)$
$△ H o = 122.56 K J$
We know $△ H = △ U + △ n R T$
$△ U = △ H - △ n R T$
$△ n = 1, R = 8.314 \times 10^{-} 3 k J m o l^{-} 1 K^{-} 1, T = 298 K$
$△ U = - 122.56 - (1) (8.314 \times 10^{-} 3) (298)$
$△ U = - 122.56 - 2.477$
$△ U = - 125.037 k J$

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The molar heat of formation of NH4 NO3(s) is −367.5KJ and those of N2O(g) and H2O(l) are +81.46KJ and −285.78KJ respectively at 25o C and 1 atmospheric pressure. The △ U for the reaction NH4NO3(s)→N2O(g) + 2H2O(l)

The molar heat of formation of $N H_{4} N O_{3} (s)$ is $- 367.5 K J$ and those of $N_{2} O (g)$ and $H_{2} O (l)$ are $+ 81.46 K J$ and $- 285.78 K J$ respectively at $25^{o} C$ and 1 atmospheric pressure. The $△$ U for the reaction
$N H_{4} N O_{3} (s) \to N_{2} O (g) + 2 H_{2} O (l)$