One gm sample of NH 4 NO 3 ( s ) is decomposed in a bomb calorimeter (constant volume). The temperature of the calorimeter system rises by 6 K . If the heat capacity of system is 1.25 kJ / K , wha

Question

One gm sample of NH 4 ​ NO 3 ​ ( s ) is decomposed in a bomb calorimeter (constant volume). The temperature of the calorimeter system rises by 6 K . If the heat capacity of system is 1.25 kJ / K , what is the molar enthalpy of decomposition of NH 4 ​ NO 3 ​ ( s ) at 300 K . [ R = 8.3 JK − 1 mol − 1 ]

TestHub · Accepted Answer

Here's a short solution to the problem:

First, calculate the heat absorbed by the calorimeter:

$q = C Δ T = (1.25 k J / K) (6 K) = 7.5 k J$

This heat is released by the decomposition of 1 gram of NH₄NO₃. Hence,

the number of moles of NH₄NO₃ in 1 gram:

$n = \frac{1 g}{80 g / m o l} = 0.0125 m o l$

The molar enthalpy of decomposition is the heat released per mole:

$Δ H = - \frac{q}{n} = - \frac{7.5 k J}{0.0125 m o l} = - 600 k J / m o l$

However, since the reaction is carried out at constant volume, we need to correct for the change in volume to obtain the enthalpy change. The decomposition reaction is:

$N H_{4} N O_{3} (s) \to N_{2} O (g) + 2 H_{2} O (g)$

The change in the number of moles of gas is $Δ n_{g} = 3 - 0 = 3$ .

We can use the following equation to relate $Δ H$ and $Δ U$ :

$Δ H = Δ U + Δ n_{g} RT$

Here, $Δ U = - 600 k J / m o l$ .

$Δ H = - 600 \times 1 0^{3} J / m o l + (3) (8.3 J / m o l K) (300 K)$

$Δ H = - 600000 + 7482 = - 592518 J / m o l = - 592.518 k J / m o l$

Chemistry - Thermochemistry Question with Solution | TestHub

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