Question

За термохимическим уравнением3H2 (г.) + N2 (г.) = 2NH3 (г.); ∆H = –46,2 кДжвычислите количество (кДж) тепловой энергии, если в результате реакции выделился амониак объемом 224 мл (н. у.)

Answer 1

 224 мл = 0,224л

                  0,224 л               х
3H₂ + N₂ = 2NH₃ (г.); ∆H = –46,2 кДж
                2*22,4                 46,2

$\frac{0.224}{2*22.4} = \frac{x}{46.2} \\ \\ x= \frac{46.2*0.224}{2*22.4} =0.231$

Ответ: Q = 0.231 кДж.

Answer 2

Calculation of Heat Energy Released in the Reaction

The given thermochemical equation is:

3H2(g) + N2(g) → 2NH3(g) (∆H = -46.2 kJ)

To calculate the amount of heat energy released in the reaction, we need to know the volume of ammonia (NH3) produced. According to the question, the volume of ammonia produced is 224 mL at standard conditions (n.u.).

To calculate the heat energy released, we can use the equation:

q = ∆H * n

where q is the heat energy released, ∆H is the enthalpy change per mole of reaction, and n is the number of moles of the substance involved in the reaction.

First, let's calculate the number of moles of ammonia produced:

Using the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At standard conditions (n.u.), the pressure is 1 atm and the temperature is 273 K.

Converting the volume of ammonia from mL to L:

224 mL = 0.224 L

Using the ideal gas law equation:

(1 atm) * (0.224 L) = n * (0.0821 L·atm/mol·K) * (273 K)

n = (1 atm * 0.224 L) / (0.0821 L·atm/mol·K * 273 K)

n = 0.0094 mol

Now, let's calculate the heat energy released:

q = ∆H * n

q = -46.2 kJ/mol * 0.0094 mol

q ≈ -0.433 kJ

Therefore, the amount of heat energy released in the reaction is approximately -0.433 kJ.