Chemistry, 03.07.2019 12:30 adrielalvarez267

Given the following data: n2(g) + 4h2(g) + cl2(g) → 2nh4cl(s) δh = -628.9 kj n2(g) + 3h2(g) → 2nh3(g) δh = -92.2 kj 2hcl(g) → h2(g) + cl2(g) δh = 184.7 kj find the δh of the following reaction: nh4cl(s) → nh3(g) + hcl(g)

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The heat of a reaction is the required amount of heat that must be added or removed from a reaction, which ids given by the enthalpy change of the reaction at constant temperature and pressure

The ΔH of the reaction NH₄Cl(s) → NH₃(g) + HCl(g) is 176 kJ

The reason why the ΔH value is correct is given as follows:

The given chemical reaction can be presented as follows;

NH₄Cl(s) → NH₃(g) + HCl(g)

The given heat of formation are given as follows;

N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH = -628.9 kJ

N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92.5 kJ

2HCl(g) → H₂(g) + Cl₂(g) ΔH = 184.7 kJ

Required:

To find ΔH for the reaction NH₄Cl(s) → NH₃(g) + HCl(g)

In the given reaction:

The number of moles of NH₄Cl = 1 mole, ΔH = -628.9/2 kJ

Number of moles of NH₃ = 1 mole, ΔH = -92.2/2 kJ

Number of moles of HCl = 1 mole, ΔH = 184.7/2 kJ

Heat of rxn = Heat of formation of products - Heat of formation of reactants

ΔHrxn = -92.35 kJ - 46.1 kJ -(-314.45 kJ) = 176 kJ

The heat of formation of the reaction, ΔHrxn = 176 kJ

Learn more about heat of a reaction here:

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Answer from: Itslegit3

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let , , and be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$ $\text{NH}_3\; (g)$ : $-2 \; b = -1$ $\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

Answer from: Quest

if you mean hydrogen and oxygen then water is the product

Answer from: Quest

The answer is 1.82 j/g°c

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