Question

NH$_4$NO$_3$(aq) $\rightarrow$ N$_2$O(g) + 2H$_2$O(l) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.72 moles of NH$_4$NO$_3$(aq) react at standard conditions. Substance S$^o$ $\left(\frac{J}{K \cdot mol}\right)$ N$_2$O(g) 219.9 H$_2$O(l) 69.9 NH$_4$NO$_3$(aq) 259.8 $\Delta S^o_{system} = \boxed{}$ J/K

          NH$_4$NO$_3$(aq) $\rightarrow$ N$_2$O(g) + 2H$_2$O(l)
Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.72 moles
of NH$_4$NO$_3$(aq) react at standard conditions.
Substance S$^o$  $\left(\frac{J}{K \cdot mol}\right)$
N$_2$O(g)     219.9
H$_2$O(l)     69.9
NH$_4$NO$_3$(aq) 259.8
$\Delta S^o_{system} = \boxed{}$ J/K

NH4NO3(aq) → N2O(g) + 2H2O(l)
Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.72 moles
of NH4NO3(aq) react at standard conditions.
Substance S^o ((J)/(K · mol))
N2O(g) 219.9
H2O(l) 69.9
NH4NO3(aq) 259.8
Δ S^osystem = J/K

Added by Austin I.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

06/11/2023

Step 1

The balanced chemical equation for the reaction is: NH4NO3(aq) -> NO(g) + 2H2O(l) From the equation, we can see that 1 mole of NH4NO3 produces 1 mole of NO and 2 moles of H2O. The change in entropy for the reaction can be calculated using the formula: ΔS = Show more…

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NH4NO3aq-NOg+2HOl) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.72moles of NH4NO(aq)react at standard conditions. J Substance So K.mol NO(g) HO(l) NH4NO(aq) 219.9 69.9 259.8 So system J/K