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TEA [102]
3 years ago
14

stoichiometric amounts of nitrogen gas and hydrogen gas react in a calorimeter to produce 5.00 g of ammonia gas. the calorimeter

temperature rises 0.42°c. the calorimeter and water have a combined heat capacity of 32.16 kj/k. calculate the heat of formation of ammonia, δhf°, in kj/mol. the formation reaction for ammonia is: 0.5n2(g) 1.5h2(g) → nh3(g)
Chemistry
2 answers:
mario62 [17]3 years ago
6 0
First, we determine the energy released by the reaction using the heat capacity and change in temperature as such:

Q = cΔT
Q = 32.16 * 0.42
Q = 13.51 kJ

Next, we determine the moles of ammonia formed as the heat of formation is expressed in "per mole".

Moles = mass / molecular weight
Moles = 5/17
Moles = 0.294

Heat of formation = 13.51 / 0.294

The heat of formation of ammonia is 45.95 kJ/mol
agasfer [191]3 years ago
4 0

The heat of formation of ammonia is \boxed{46.02{\text{ kJ/mol}}}.

Further Explanation:

Standard enthalpy of formation:

It is the change in enthalpy when one mole of the substance is formed from its constituent elements. It is represented by \Delta H_{\text{f}}^\circ.

The given reaction occurs as follows:

{\text{0}}{\text{.5}}{{\text{N}}_2}\left(g\right)+1.5{{\text{H}}_2}\left(g\right)\to{\text{N}}{{\text{H}}_{\text{3}}}\left(g\right)

The formula to calculate the moles of {\text{N}}{{\text{H}}_{\text{3}}} is as follows:

{\text{Moles of N}}{{\text{H}}_{\text{3}}} = \frac{{{\text{Given mass of N}}{{\text{H}}_{\text{3}}}}}{{{\text{Molar mass of N}}{{\text{H}}_{\text{3}}}}}             ...... (1)

The given mass of {\text{N}}{{\text{H}}_{\text{3}}} is 5 g.

The molar mass of {\text{N}}{{\text{H}}_{\text{3}}} is 17.03 g/mol.

Substitute these values in equation (1).

\begin{aligned}{\text{Moles of N}}{{\text{H}}_{\text{3}}}&=\left({5{\text{ g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{17}}{\text{.03 g}}}}}\right)\\&=0.2935{\text{ mol}}\\\end{aligned}

The formula to calculate the heat energy of reaction is as follows:

{\text{Q}}={c\Delta T}}                                        …… (2)

Here,

Q is the amount of heat transferred.

c is the specific heat of the substance.

{\Delta T}} is the change in temperature of substance.

The temperature change is to be converted into K. The conversion factor for this is,

\Delta{\text{T}}\left({^\circ{\text{C}}}\right)=\Delta{\text{T}}\left({\text{K}}\right)

So the value of {\Delta T}} becomes 0.42 K.

The value of c is 32.16 kJ/K.

The value of {\Delta T}} is 0.42 K.

Substitute these values in equation (2).

\begin{aligned}{\text{Q}}&=\left({{\text{32}}{\text{.16 kJ/K}}}\right)\left({{\text{0}}{\text{.42}}\;{\text{K}}}\right)\\&=13.50{\text{72 kJ}}\\\end{aligned}

The heat of formation of ammonia is calculated as follows:

\begin{aligned}\Delta H_{\text{f}}^\circ&=\frac{{{\text{13}}{\text{.5072 kJ}}}}{{0.2935{\text{ mol}}}}\\&=46.0211{\text{ kJ/mol}}\\&\approx{\text{46}}{\text{.02 kJ/mol}}\\\end{aligned}

Learn more:

1. What is the final temperature of copper? brainly.com/question/7210400

2. Calculate the specific heat of gold: brainly.com/question/3032746

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: Q, c, 0.42 K, 13.5072 kJ, 46.02 kJ/mol, ammonia, specific heat, heat of formation.

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