/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 125 Heavy water, symbolized \(\mathr... [FREE SOLUTION] | 91Ó°ÊÓ

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Heavy water, symbolized \(\mathrm{D}_{2} \mathrm{O}\left(\mathrm{D}={ }^{2} \mathrm{H}\right)\) finds use as a neutron moderator in nuclear reactors. In a mixture with ordinary water, exchange of isotopes occurs according to the following equation: $$ \mathrm{H}_{2} \mathrm{O}+\mathrm{D}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{HDO} \quad K_{c}=3.86 \text { at } 298 \mathrm{~K} $$ When \(1.00\) mol of \(\mathrm{H}_{2} \mathrm{O}\) is combined with \(1.00 \mathrm{~mol}\) of \(\mathrm{D}_{2} \mathrm{O}\), what are the equilibrium amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{D}_{2} \mathrm{O}\), and \(\mathrm{HDO}(\) in moles \()\) at 298 \(\mathrm{K}\) ? Assume the density of the mixture is constant at \(1.05 \mathrm{~g} / \mathrm{cm}^{3}\).

Short Answer

Expert verified
Equilibrium amounts: \( [\mathrm{H}_{2} \mathrm{O}] = 0.337 \) mol, \( [\mathrm{D}_{2} \mathrm{O}] = 0.337 \) mol, \( [\mathrm{HDO}] = 1.326 \) mol.

Step by step solution

01

Setup the reaction equation

The reaction we're analyzing is given as \( \mathrm{H}_{2} \mathrm{O} + \mathrm{D}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{HDO} \).
02

Define initial amounts

The initial amounts are \( 1.00 \) mol of \( \mathrm{H}_{2} \mathrm{O} \) (\( x_0 = 1.00 \) mol) and \( 1.00 \) mol of \( \mathrm{D}_{2} \mathrm{O} \) (\( y_0 = 1.00 \) mol). The initial amount of \( \mathrm{HDO} \) is \( 0 \) mol.
03

Define change in amounts

Let \( x \) be the amount of \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{D}_{2} \mathrm{O} \) that reacts to form \( 2x \) mol of \( \mathrm{HDO} \). The changes are \( -x \) for both \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{D}_{2} \mathrm{O} \), and \( +2x \) for \( \mathrm{HDO} \).
04

Express equilibrium amounts

The equilibrium amounts are \( \mathrm{H}_{2} \mathrm{O} = 1.00 - x \), \( \mathrm{D}_{2} \mathrm{O} = 1.00 - x \), and \( \mathrm{HDO} = 2x \).
05

Write the expression for Kc

The equilibrium constant expression is \( K_c = \frac{[\mathrm{HDO}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{D}_{2} \mathrm{O}]} \). Substituting in the equilibrium amounts gives \( K_c = \frac{(2x)^2}{(1.00-x)^2} \).
06

Solve for x using Kc value

Given that \( K_c = 3.86 \), we have \( \frac{4x^2}{(1.00-x)^2} = 3.86 \). Solving this quadratic equation (\( \sqrt{3.86} = 1.964 \)) gives \( \frac{2x}{1.00 - x} = 1.964 \). Solving further, \( 2x = 1.964 - 1.964x \). Simplifying, \( 2.964x = 1.964 \), thus \( x \approx 0.663 \).
07

Calculate equilibrium amounts

Substitute \( x = 0.663 \) into the equilibrium expressions: \( [\mathrm{H}_{2} \mathrm{O}] = 1.00 - 0.663 = 0.337 \) mol, \( [\mathrm{D}_{2} \mathrm{O}] = 1.00 - 0.663 = 0.337 \) mol, \( [\mathrm{HDO}] = 2(0.663) = 1.326 \) mol.

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Key Concepts

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

Isotope Exchange
When two isotopes interact, a phenomenon known as isotope exchange can occur. In the context of our exercise, we're dealing with hydrogen isotopes in heavy water (D₂O) and ordinary water (H₂O). Heavy water contains deuterium, which is a heavier isotope of hydrogen, denoted as ²H. When these isotopes come in contact, an exchange happens between hydrogen atoms. For the reaction \( \mathrm{H}_2\mathrm{O} + \mathrm{D}_2\mathrm{O} \rightleftharpoons 2 \mathrm{HDO} \), the isotopes are exchanging places to form a new compound, HDO.
  • Hâ‚‚O splits into two hydrogen atoms and an oxygen atom.
  • Similarly, Dâ‚‚O splits into two deuterium atoms and an oxygen atom.
  • These atoms then recombine to form the mixed isotope product, HDO.
This reaction is dynamic, meaning it can go forward or backward until it reaches a point where the rate of exchange in both directions is balanced, known as equilibrium.
Equilibrium Constant
In chemical reactions, like our isotope exchange, the equilibrium constant, \(K_c\), is crucial. The equilibrium constant tells us how much of each substance is present at equilibrium. It's a ratio of the concentration of the products to the reactants raised to the power of their coefficients in the balanced reaction equation. For the reaction \( \mathrm{H}_{2} \mathrm{O} + \mathrm{D}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{HDO} \), the expression for \( K_c \) is:\[K_c = \frac{[\mathrm{HDO}]^2}{[\mathrm{H}_{2} \mathrm{O}][\mathrm{D}_{2} \mathrm{O}]}\]Here, \([\mathrm{HDO}]\) is the concentration of the product and \([\mathrm{H}_{2} \mathrm{O}]\) and \([\mathrm{D}_{2} \mathrm{O}]\) are the concentrations of the reactants at equilibrium. A higher \( K_c \) value, like 3.86 in our example, suggests that the reaction favors the products (HDO in this case). However, if \( K_c \) were less than 1, the reactants would dominate. Equilibrium is not about equal amounts but rather a point of balance in terms of concentration.
Nuclear Chemistry
This exercise also relates to nuclear chemistry due to the involvement of isotopes. In nuclear chemistry, isotopes of an element can have vastly different properties. For instance, deuterium (²H) is a stable isotope of hydrogen with one neutron, distinguishing it from the common hydrogen isotope, which has no neutrons. Heavy water \(\mathrm{D}_{2} \mathrm{O}\) is particularly important in nuclear reactors because:
  • It acts as a neutron moderator, slowing down neutrons to sustain a controlled nuclear chain reaction.
  • Due to its increased mass, deuterium is less likely to absorb neutrons compared to ordinary hydrogen.
Understanding isotopes and their roles in reactions like these is central to exploring nuclear chemistry applications. The ability of heavy water to facilitate nuclear reactions without becoming radioactive itself makes it invaluable in this field.

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Most popular questions from this chapter

For each of the following equilibria, write the equilibrium constant expression for \(K_{c}\). Where appropriate, also write the equilibrium constant expression for \(K_{\mathrm{p}^{-}}\) (a) \(\mathrm{WO}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{W}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s)\) (c) \(2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(g)+6 \mathrm{HCl}(g)\) (d) \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{CO}_{2}(g)\)

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