Question

What is the enthalpy change (in kj) of a chemical reaction that raises the temperature of 250.0 ml of solution having a density of 1.25 g/ml by 9.20 ∘c? (the specific heat of the solution is 3.74 j/g⋅k.)?

Answer 1

The enthalpy change (in kJ) for the given solution is $\boxed{{\text{329}}{\text{.82 kJ}}}$

Further explanation:

The property is a unique feature of the substance that differentiates it from the other substances. It is classified into two types:

1. Intensive properties:

These are the properties that depend on the nature of the substance. These don't depend on the size of the system. Their values remain unaltered even if the system is further divided into a number of subsystems. Temperature, refractive index, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

These are the properties that depend on the amount of the substance. These are additive in nature when a single system is divided into many subsystems. Mass, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.

Enthalpy:

It is a thermodynamic property that is defined as the sum of internal energy and product of pressure (P) and volume (V) of the system. It is a state function, an extensive property, and is independent of the path followed by the system while moving from initial to the final point. The total enthalpy of the system cannot be measured directly so its change $\left({\Delta{\text{H}}}\right)$ is usually measured.

The enthalpy change $\left({\Delta{\text{H}}}\right)$can have two values:

Case I: If the reaction is endothermic, more energy needs to be supplied to the system than that released by it. So $\Delta{\text{H}}$ comes out to be positive.

Case II: If the reaction is exothermic, more energy is released by the system than that supplied to it. So $\Delta{\text{H}}$ comes out to be negative.

Specific heat is the amount of heat required to increase the temperature of any substance per unit mass. Specific heat capacity is also known as or mass specific heat. Its SI unit is Joule (J).

The formula to calculate the heat energy of any substance is as follows:

${\text{Q}}={mc\Delta T}}$                       …… (1)

Here,

Q is the amount of heat transferred.

m is the mass of the substance.

c is the specific heat of the substance.

${\Delta T}}$ is the change in temperature of the system.

The formula to calculate the density of the solution is as follows:

${\text{Density of solution}}=\frac{{{\text{Mass of solution}}}}{{{\text{Volume of solution}}}}$               ….. (2)

Rearrange equation (2) for the mass of the solution.

${\text{Mass of solution}}=\left({{\text{Density of solution}}}\right)\left({{\text{Volume of solution}}}\right)$        …… (3)

The density of the solution is 1.25 g/mL.

The volume of solution is 250 mL.

Substitute these values in equation (3).

$\begin{gathered}{\text{Mass of solution}}=\left({\frac{{{\text{1}}{\text{.25 g}}}}{{1\;{\text{mL}}}}}\right)\left({{\text{250 mL}}}\right)\\={\text{312}}{\text{.5 g}}\\\end{gathered}$

The temperature change $\left({\Delta{\text{T}}}\right)$ is to be converted to K. The conversion factor for this is,

${\text{0 }}^\circ{\text{C}}={\text{273 K}}$

So ${\Delta T}}$ can be calculated as follows:

$\begin{gathered}{\text{Temperature}}\left({\text{K}}\right)=\left( {9.2+273}\right)\;{\text{K}}\\=282.2\;{\text{K}}\\\end{gathered}$

The mass of the solution is 312.5 g.

The specific heat of the solution is $3.74\;{\text{J/g K}}$.

${\Delta T}}$ of the system is 282.2 K.

Substitute these values in equation (1).

$\begin{gathered}{\text{Q}}=\left({{\text{312}}{\text{.5 g}}}\right)\left({\frac{{3.74\;{\text{J}}}}{{\left({{\text{1 g}}}\right)\left({{\text{1 K}}}\right)}}}\right)\left({282.2\;{\text{K}}}\right)\\=329821.25\;{\text{J}}\\\end{gathered}$

The enthalpy change is to be converted into kJ. The conversion factor for this is,

${\text{1 J}}={10^{-3}}\;{\text{kJ}}$

So the enthalpy change can be calculated as follows:

$\begin{gathered}{\text{Q}}=\left({329821.25\;{\text{J}}}\right)\left({\frac{{{{10}^{-3}}\;{\text{kJ}}}}{{{\text{1 J}}}}}\right)\\=329.82125\;{\text{kJ}}\\\approx{\text{329}}{\text{.82 kJ}}\\\end{gathered}$

Therefore, the enthalpy change of the given reaction is 329.82 kJ.

Learn more:

1. What is the equilibrium constant of pure water at  ? brainly.com/question/3467841

2. 1. Calculate   for the reaction using Hess law: brainly.com/question/11293201

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: intensive, extensive, enthalpy, mass of solution, amount of heat transferred, Q, m, c, given mass, molar mass, enthalpy change, 329.82 kJ, enthalpy change, density of solution, mass of solution, volume of solution, conversion factor, 250 mL, 1.25 g/mL.

Answer 2

250 ml * 1.25 g/ml * 3.74 j/g-K * 9.2 K = 10.752 kJ Pretty much, all you need to do here is multiply all of these out to get your final answer. Not all questions are this easy, but this is certainly one of them.