When 10.0 g of ammonium nitrate is dissolved in 100 cm3 of water, the temperature of the solution decreases from 19.0°C to 10.5°C, how do I solve the following?

Question

a) Compare the total enthalpy of the reactants with that of the products and state which is greater.
b) Classify this reaction as endothermic or exothermic.
c) State which chemical bonds were broken when the ammonium nitrate
dissolved in the water.

Luna Chen · Accepted Answer

(a) #ΔH_text(products) > ΔH_text(reactants)#; (b) endothermic; (c) ionic bonds between #"NH"_4^+"# and #"NO"_3^"-"#.

(a) Enthalpy of formation of reactants and products We have the reaction #M_r"/g·mol"^"-1"color(white)(mmmll)80.04# #color(white)(mmmmmmmm)"NH"_4"NO"_3"(s)" ⇌ "NH"_4^"+""(aq)" + "NO"_3^"-""(aq)"# #Δ_fH^°"/kJ·mol"^"-1"color(white)(m)"-365.1"color(white)(mmmm)"-132.8"color(white)(mmm)"-206.6"# #Δ_fH_text(reactants)^° = "-365.1 kJ.mol"# #Δ_fH_text(products)^° = "(-132.8 - 206.6) kJ" = "-339.4 kJ/mol"# #"Moles of NH"_4"NO"_3 = 10.0 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × ("1 mol NH"_4"NO"_3)/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3)))) = "0.1249 mol NH"_4"NO"_3 = "0.1249 mol reaction"# #ΔH_text(reactants) = 0.1249 color(red)(cancel(color(black)("mol reaction"))) × "-365.1 kJ"/(1 color(red)(cancel(color(black)("mol reaction")))) = "-45.6 kJ"# #ΔH_text(products) = 0.1249 color(red)(cancel(color(black)("mol reaction"))) ×"-339.4 kJ"/(1 color(red)(cancel(color(black)("mol reaction")))) = "-42.4 kJ"# ∴ #ΔH_text(products) > ΔH_text(reactants)# (b) Exothermic or endothermic? The temperature decreased when the ammonium nitrate dissolved in the water. Energy left the system, so the reaction is endothermic. (c) Bonds broken The ionic bonds between the #"NH"_4^+"# ions and #"NO"_3^"-"# ions were broken when the ammonium nitrate dissolved in the water.

Naomi Aronson · Answer

undefined Calculate the heat absorbed using the formula: $ q = m \cdot c \cdot \Delta T $, where $ m = 10.0 \, \text{g} $, $ c = 4.18 \, \text{J/g°C} $, $ \Delta T = (10.5 - 19.0) \, \text{°C} $.

Serenity Allen · Answer

undefined To solve this, you need to use the formula for calculating the heat absorbed or released during a change in temperature. The formula is:

Q = m * c * ΔT

where:
Q = heat absorbed or released (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (final temperature - initial temperature) (in °C)

First, you need to calculate the heat absorbed or released when 10.0 g of ammonium nitrate is dissolved in water. The specific heat capacity of water is 4.18 J/g°C.

1. Calculate the change in temperature (ΔT):
ΔT = 10.5°C - 19.0°C
ΔT = -8.5°C

2. Calculate the heat absorbed or released (Q):
Q = 10.0 g * 4.18 J/g°C * -8.5°C

The negative sign indicates that heat is released by the solution as it cools down. The final step is to calculate the amount of heat released in joules.

When 10.0 g of ammonium nitrate is dissolved in 100 cm3 of water, the temperature of the solution decreases from 19.0°C to 10.5°C, how do I solve the following?

a) Compare the total enthalpy of the reactants with that of the products and state which is greater. b) Classify this reaction as endothermic or exothermic. c) State which chemical bonds were broken when the ammonium nitrate dissolved in the water.

a) Compare the total enthalpy of the reactants with that of the products and state which is greater.
b) Classify this reaction as endothermic or exothermic.
c) State which chemical bonds were broken when the ammonium nitrate
dissolved in the water.