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What is the entropy change (in \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ) when 1 mol of ice is converted into water at \(0^{\circ} \mathrm{C}\) ? (The enthalpy change for the conversion of ice to liquid water is \(6.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(0^{\circ} \mathrm{C}\) ) a. \(2.198\) b. \(21.98\) c. \(20.13\) d. \(2.013\)

Short Answer

Expert verified
The entropy change is option b: 21.98 \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Step by step solution

01

Formula for Entropy Change

To determine the change in entropy for the transition of ice to liquid water, we will use the formula: \[ \Delta S = \frac{\Delta H}{T} \] where \( \Delta S \) is the change in entropy, \( \Delta H \) is the change in enthalpy, and \( T \) is the temperature in Kelvin.
02

Convert Temperature to Kelvin

The process occurs at \( 0^{\circ} \mathrm{C} \), which needs to be converted to Kelvin. Remember, \( T(\mathrm{K}) = T(\mathrm{C}) + 273.15 \). Thus, \( 0^{\circ} \mathrm{C} \) is equivalent to \( 273.15 \mathrm{K} \).
03

Plug Values into the Formula

We have \( \Delta H = 6.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \), which must be converted to \( \mathrm{J} \mathrm{~mol}^{-1} \). So, \( 6.0 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \). Now, substitute \( \Delta H = 6000 \mathrm{~J} \mathrm{~mol}^{-1} \) and \( T = 273.15 \mathrm{K} \) into the formula to find \( \Delta S \): \[ \Delta S = \frac{6000}{273.15} \approx 21.98 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \].
04

Select Correct Answer Option

From our calculation, the entropy change is \( 21.98 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \), which corresponds to option b.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. It's denoted by the symbol \( \Delta H \). Enthalpy is a fundamental concept in understanding how energy transfers in reactions and physical processes. When ice melts into water, it absorbs heat from its surroundings. This is why the enthalpy change is positive for this transition. The enthalpy change for converting 1 mole of ice to water is provided as \( 6.0 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \). This value indicates the amount of energy required in kilojoules to melt 1 mole of ice at its melting point of \( 0^{\circ} \mathrm{C} \).
Remember, in calculations, it's crucial to convert enthalpy values to joules, since standard entropy units are \( \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} \). To do this, multiply by \( 10^3 \) because \( 1 \, \mathrm{kJ} = 1000 \, \mathrm{J} \). In our example, the enthalpy change becomes \( 6000 \, \mathrm{J} \, \mathrm{mol}^{-1} \).
Phase Transition
A phase transition is when a substance changes from one state of matter to another, such as solid to liquid. In the exercise, we're looking at ice changing to water, which is a common example of a phase transition.
  • During this change, molecules gain enough energy to break free from their rigid structure of a solid.
  • This process occurs at a specific temperature, the melting point, for ice it is \(0^{\circ} \mathrm{C}\).

The concept of entropy, which is the measure of disorder or randomness in a system, tends to increase as substances transition from an orderly solid to a more disorderly liquid.
It's crucial to use the formula \( \Delta S = \frac{\Delta H}{T} \) to quantify this entropy change. Here, \( \Delta H \) signifies the enthalpy change associated with the transition, while \( T \) is the temperature in Kelvin at which the phase transition occurs.
Kelvin Temperature Conversion
Temperature conversion from Celsius to Kelvin is essential when applying thermodynamic formulas, which use Kelvin (K) as the temperature unit. Kelvin is an absolute temperature scale, that starts at absolute zero, the theoretically coldest possible temperature.

Converting Celsius to Kelvin is straightforward:
  • Add 273.15 to the Celsius temperature. For example, at \( 0^{\circ} \mathrm{C} \), the Kelvin equivalent is \( 273.15 \, \mathrm{K} \).

This conversion ensures that you are using consistent units within equations, as required in the formula for calculating entropy change, \( \Delta S = \frac{\Delta H}{T} \). Using Kelvin in calculations prevents confusion because negative temperatures in thermodynamics would not be physically meaningful.

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Most popular questions from this chapter

For the reaction, \(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}, \Delta \mathrm{H}=-560 \mathrm{~kJ}\). Two moles of \(\mathrm{CO}\) and one mole of \(\mathrm{O}_{2}\) are taken in a container of volume \(1 \mathrm{~L}\). They completely form two moles of \(\mathrm{CO}_{2}\), the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to \(40 \mathrm{~atm}\), find the magnitude (absolute value) of \(\Delta \mathrm{U}\) at \(500 \mathrm{~K}\). (1 L atm \(=0.1 \mathrm{~kJ})\) a. \(557 \mathrm{KJ}\) b. \(55.7 \mathrm{KJ}\) c. \(278 \mathrm{KJ}\) d. \(27.8 \mathrm{KJ}\) [IIT 2006]

(A): \(\Delta \mathrm{E}\) is state function of the system. (R): As it depends upon the final and initial state of the system.

Which of the following reaction defines \(\Delta \mathrm{H}^{\circ}\) ? a. \(\mathrm{C}\) (diamond) \(+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) b. \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g})\) c. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) d. \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) [IIT 2003]

(A): There is a natural asymmetry between converting work to heat and converting heat to work. (R): No process is possible in which the sole result in the absorption of heat from a reservation and its complete conversion into work

In certain chemical process both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) have values greater than zero. Under what conditions, the reaction would not be spontaneous? a. \(\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}\) b. \(\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S}\) c. \(\Delta \mathrm{G}>0\) d. \(\Delta \mathrm{G}<0\)

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