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Chapter 15: Problem 41

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?

Short Answer

Expert verified
The total pressure at equilibrium under given conditions is approximately 5.61 atm

Step by step solution

01

Understanding the information

Formamide (HCONH2) decomposes into NH3 and CO. It is given that initially, 0.186 mol of HCONH2 was present and it dissociates at 400 K in a 2.16 L flask. At equilibrium, the equilibrium constant (Kc) is 4.84.
02

Set up the equilibrium table for equilibrium concentrations

At start, the moles of HCONH2 equals 0.186 mol. NH3 and CO are zero. At equilibrium, the amount of HCONH2 that dissociates can be represented as x. So, at equilibrium the amounts of gases will be: HCONH2, NH3 and CO will be 0.186 - x, x and x mol respectively.
03

Substitute given quantities into equilibrium constant expression

The expression for Kc is [NH3][CO]/[HCONH2]. Substituting values we get: 4.84=(x/2.16)*(x/2.16)/((0.186-x)/2.16). Simplifying this equation will allow us to solve for x.
04

Solve for x

Solving the equation, we get x approximately equal to 0.115 mol. This means, at equilibrium, we have 0.115 mol of NH3 and CO each and 0.071 mol of HCONH2.
05

Calculate total pressure at equilibrium

Total pressure can be calculated using the Ideal Gas Law: P = nRT/V. In this case, n is total number of moles of all gases, R is the gas constant (0.0821 L·atm/K·mol), T is the temperature in Kelvin and V is the volume in liters. Substituting for n = 0.115 (NH3) + 0.115 (CO) + 0.071 (HCONH2) = 0.301 mol, T = 400 K, and V = 2.16 L, the equation yields the total pressure at equilibrium.

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Most popular questions from this chapter

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

An important environmental and physiological reaction is the formation of carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) from carbon dioxide and water. Write the equilibrium constant expression for this reaction in terms of activities. Convert that expression into an equilibrium constant expression containing concentrations and pressures.

Determine values of \(K_{c}\) from the \(K_{p}\) values given. (a) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=2.9 \times 10^{-2} \mathrm{at} 303 \mathrm{K}\) (b) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=1.48 \times 10^{4} \mathrm{at} 184^{\circ} \mathrm{C}\) (c) \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(K_{\mathrm{p}}=0.429\) at \(713 \mathrm{K}\)

Based on these descriptions, write a balanced equation and the corresponding \(K_{c}\) expression for each reversible reaction. (a) Carbonyl fluoride, \(\mathrm{COF}_{2}(\mathrm{g}),\) decomposes into gaseous carbon dioxide and gaseous carbon tetrafluoride. (b) Copper metal displaces silver(I) ion from aqueous solution, producing silver metal and an aqueous solution of copper(II) ion. (c) Peroxodisulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\), oxidizes iron(II) ion to iron(III) ion in aqueous solution and is itself reduced to sulfate ion.

A 1.100 L flask at \(25^{\circ} \mathrm{C}\) and 1.00 atm pressure contains \(\mathrm{CO}_{2}(\mathrm{g})\) in contact with \(100.0 \mathrm{mL}\) of a saturated aqueous solution in which \(\left[\mathrm{CO}_{2}(\mathrm{aq})\right]=3.29 \times 10^{-2} \mathrm{M}\) (a) What is the value of \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq}) ?\) (b) If 0.01000 mol of radioactive \(^{14} \mathrm{CO}_{2}\) is added to the flask, how many moles of the \(^{14} \mathrm{CO}_{2}\) will be found in the gas phase and in the aqueous solution when equilibrium is re-established? [Hint: The radioactive \(^{14} \mathrm{CO}_{2}\) distributes itself between the two phases in exactly the same manner as the nonradioactive \(\left.^{12} \mathrm{CO}_{2} .\right]\)
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