Question

Question 1:

A compound A reacts according to the following hypothetical equation and has a molecular weight of 48.36 g/mol.

3 A (s) +B (aq) ––> 2 C (aq) ∆H° = ?

A sample of A, weighing 0.152 g reacts in a flask containing 250.00 g of water and the water temperature increases from 24.85 °C to a temperature of 26.26 °C. Calculate ∆H° for the reaction as written in the equation.

Answer: -1408 kJ

Question 2:

For which of the following reactions is ∆H° = ∆H°f, the heat of formation?
i. C (s) + 2 F2(g) ––> CF4 (g) ∆H° = – 221.0 kJ
ii. H(g) + Br (g) ––> HBr (g) ∆H° = –366.2 kJ
iii. 2 C(s) + H2(g) + 3 Cl2(g) ––> 2 CHCl3 (g) ∆H° = –268.2

Could you also explain exactly what Delta H means.

Answer 1

Answer on Question #49936, Chemistry, Inorganic Chemistry

Question 1:

A compound A reacts according to the following hypothetical equation and has a molecular weight of 48.36 g/mol.

3 A (s) + B (aq) → 2 C (aq) ΔH° = ?

A sample of A, weighing 0.152 g reacts in a flask containing 250.00 g of water and the water temperature increases from 24.85 °C to a temperature of 26.26 °C. Calculate ΔH° for the reaction as written in the equation.

Answer: -1408 kJ

Question 2:

For which of the following reactions is ΔH° = ΔH°ₜ, the heat of formation?

i. C (s) + 2 F₂(g) → CF₄ (g) ΔH° = -221.0 kJ

ii. H(g) + Br (g) → HBr (g) ΔH° = -366.2 kJ

iii. 2 C(s) + H₂(g) + 3 Cl₂(g) → 2 CHCl₃ (g) ΔH° = -268.2

Could you also explain exactly what Delta H means.

Solution:

Delta H (ΔH) is defined as the amount of heat evolved or absorbed in the reacting species.

Question 1:

Hearing of water:

The amount of heat spent on heating of water (Q) equal to the amount of heat that was allocated in the chemical reaction (ΔH):

$$\Delta H = Q = c_{H_2O} \times m_{H_2O} \times (T_2 - T_1)$$

$c_{H_2O}$ is a constant 4.187×10³ J/(kg×K). So:

$$\Delta H = 4.187 \times 10^3 \times 0.25 \times (299.41 - 298) = 1.476 \times 10^3 = 1.476 \, \text{kJ}$$ $$v = \frac{\Delta H}{\Delta H^0}$$ $$\Delta H^0 = \frac{\Delta H}{v}$$ $$v = \frac{v_a}{3} = \frac{m_a}{M_r \times 3}$$ $$\Delta H^0 = \frac{\Delta H \times M_r \times 3}{m_a} = \frac{1.476 \times 48.36 \times 3}{0.152} = 1408 \, \text{kJ/mol}$$ $$\Delta H^0 = \sum \Delta H_f^0 \, (\text{products}) - \sum \Delta H_f^0 \, (\text{reagents})$$

ΔHₜ⁰ is the standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy from the formation of 1 mole of the compound from its constituent elements, with all substances in their standard states at 1 atmosphere (1 atm or 101.3 kPa).

If the heat is produced, enthalpy change (ΔH⁰) would be negative (ΔH⁰<0)

Answer:

-1408 kJ

Question 2:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy from the formation of 1 mole of the compound from its constituent elements, with all substances in their standard states at 1 atmosphere (1 atm or 101.3 kPa).

Standard states are as follows:

- For a gas: the standard state is a pressure of exactly 1 atm

- For a solute present in an ideal solution: a concentration of exactly 1 M at a pressure of 1 atm

- For a pure substance or a solvent in a condensed state (a liquid or a solid): the standard state is the pure liquid or solid under a pressure of 1 atm

For the reaction iii, it's not the heat of formation because it's not formed one mole of a substance, and chloroform (CHCl₃) under normal conditions is a liquid but not a gas.

For the reaction ii, it's not the heat of formation because all substances aren't in their standard states at 1 atmosphere; should be H₂ and Br₂.

For the reaction i, it's the heat of formation, all conditions are satisfied!

Answer:

i.C (s) + 2 F2(g) → CF4 (g) ΔH° = -221.0 kJ ΔH° = ΔH°f

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