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Chapter 15: Problem 26

At \(2000 \mathrm{K}, K_{c}=0.154\) for the reaction \(2 \mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If a \(1.00 \mathrm{L}\) equilibrium mixture at \(2000 \mathrm{K}\) contains \(0.10 \mathrm{mol}\) each of \(\mathrm{CH}_{4}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (a) what is the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) present? (b) Is the conversion of \(\mathrm{CH}_{4}(\mathrm{g})\) to \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) favored at high or low pressures? (c) If the equilibrium mixture at \(2000 \mathrm{K}\) is transferred from a 1.00 L flask to a 2.00 L flask, will the number of moles of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) increase, decrease, or remain unchanged?

Short Answer

Expert verified
(a) The mole fraction of \(C2H2(g)\) present is 0.89. (b) The conversion of \(CH4(g)\) to \(C2H2(g)\) is favored at low pressures. (c) The number of moles of \(C2H2(g)\) will increase when the mixture is transferred from a 1.00L flask to a 2.00L flask.

Step by step solution

01

Understanding Given Data & Equilibrium expression

Given that at \(2000K\), \(Kc = 0.154\) for the reaction, \(2CH4(g) ⇌ C2H2(g) + 3H2(g)\) Also, we are given that 1L equilibrium mixture contains \(0.1 mol\) each of \(CH4(g)\) and \(H2(g)\). We can write equilibrium expression as \(K_c = [C_2H2][H2]^3/[CH4]^2\).
02

Calculating Concentration & Mole Fraction for Part (a)

Using the equilibrium expression: \(0.154 = [C_2H2][0.1]^3/[0.1]^2\). Therefore, \(0.154 = [C_2H2][0.1]\), which gives \([C_2H2] = 1.54 M\) which is the concentration of \(C_2H2\). Total moles of the mixture will now be \(0.1+0.1+1.54 = 1.74\), thus the mole fraction of \(C_2H2=g\) will be \(1.54/1.74 = 0.89\).
03

Understanding Pressure Impact for Part (b)

According to Le Chatelier's principle, if a system at equilibrium is disturbed, the system tends to counteract the disturbance and a shift in the position of equilibrium is observed. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas to reduce the pressure. Here, the left side of the equation has 2 moles of gas whereas the right side has 4. So, the conversion of \(CH4\) to \(C2H2\) is favored at low pressures.
04

Understanding Volume Change Impact for Part (c)

Increasing the volume is equivalent to decreasing the pressure. According to Le Chatelier's principle, this will shift the equilibrium toward the side of the reaction that has more moles of gas. Therefore, the number of moles of \(C2H2(g)\) will increase when the mixture is transferred from a 1.00L flask to a 2.00L flask.

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Most popular questions from this chapter

Write an equilibrium constant, \(K_{c},\) for the formation from its gaseous elements of \((a) 1\) mol \(\mathrm{HF}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g}) ;(\mathrm{c}) 2 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) 1 \mathrm{mol} \mathrm{ClF}_{3}(1)\)

In the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}), 0.455\) \(\mathrm{mol} \mathrm{SO}_{2}, 0.183 \mathrm{mol} \mathrm{O}_{2},\) and \(0.568 \mathrm{mol} \mathrm{SO}_{3}\) are introduced simultaneously into a 1.90 L vessel at \(1000 \mathrm{K}\). (a) If \(K_{c}=2.8 \times 10^{2},\) is this mixture at equilibrium? (b) If not, in which direction will a net change occur?

In the reversible reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{HI}(\mathrm{g}),\) an initial mixture contains \(2 \mathrm{mol} \mathrm{H}_{2}\) and 1 mol I \(_{2} .\) The amount of HI expected at equilibrium is (a) \(1 \mathrm{mol} ;\) (b) \(2 \mathrm{mol} ;\) (c) less than \(2 \mathrm{mol}\); (d) more than 2 mol but less than 4 mol.

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}),\) is caused by the oxidation of iron by oxygen. Write the equilibrium constant expression first in terms of activities, and then in terms of concentration and pressure.
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