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julsineya [31]
3 years ago
15

The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia

given the following mean bond enthalpies (kJmol-1): N≡N 942; H-H 436; N-H 386
Chemistry
1 answer:
Savatey [412]3 years ago
7 0

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

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Honors Stoichiometry Activity Worksheet
vaieri [72.5K]

Explanation: Here, we will be considering 1 cup is equal to 1 mol.

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol of water

Mass of water used =  1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with  mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol of water

Mass of sugar used =  0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol of lemonade

Mass of lemonade obtained  = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

The percentage yield of  lemonade is 95% and ingredient which remained unused were water and sugar.

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<span>B) Soap makes the water less viscous making the paperclip sink.</span>

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