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

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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] · Answer 1 · 2022-01-11T02:34:00+03:00

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] · Answer 2 · 2022-01-11T05:01:45+03:00

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:

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