Question 17 of 69intermediate🔧 ApplyNumerical3 marks

Kp=0.04K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2_2H6_6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6(g)⇌C2H4(g)+H2(g)C_2H_6(g) \rightleftharpoons C_2H_4(g) + H_2(g)
Correct Answer

The equilibrium concentration of C2_2H6_6 is 0.049 M (equilibrium partial pressure is 3.62 atm).

Exercise: EXERCISES | Q: 6.17 | (Page 42)
For More Understanding

Explanation

To find the equilibrium concentration, we first determine the equilibrium partial pressures using the given KpK_p value. By setting up the initial and equilibrium pressures, we derive a quadratic equation for the change in pressure (xx). Solving this gives the equilibrium partial pressure of C2_2H6_6, which is then converted to concentration using the ideal gas law relation C=P/RTC = P/RT.

Solution Steps

  1. Step 1: Let the initial pressure of C2H6C_2H_6 be 4.0 atm. Let xx atm of C2H6C_2H_6 decompose at equilibrium. The equilibrium partial pressures are: pC2H6=4.0−xp_{C_2H_6} = 4.0 - x, pC2H4=xp_{C_2H_4} = x, and pH2=xp_{H_2} = x.

  2. Step 2: Write the expression for KpK_p and substitute the equilibrium pressures: Kp=pC2H4×pH2pC2H6=x×x4.0−x=0.04K_p = \frac{p_{C_2H_4} \times p_{H_2}}{p_{C_2H_6}} = \frac{x \times x}{4.0 - x} = 0.04.

  3. Step 3: Rearrange the equation into a standard quadratic form: x2=0.04(4.0−x)⇒x2+0.04x−0.16=0x^2 = 0.04(4.0 - x) \Rightarrow x^2 + 0.04x - 0.16 = 0.

  4. Step 4: Solve the quadratic equation using x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1,b=0.04,c=−0.16a=1, b=0.04, c=-0.16. x=−0.04±(0.04)2−4(1)(−0.16)2=−0.04±0.64162=−0.04±0.8012x = \frac{-0.04 \pm \sqrt{(0.04)^2 - 4(1)(-0.16)}}{2} = \frac{-0.04 \pm \sqrt{0.6416}}{2} = \frac{-0.04 \pm 0.801}{2}.

  5. Step 5: The valid root is x=0.7612=0.38x = \frac{0.761}{2} = 0.38 atm (the negative root is neglected because pressure cannot be negative).

  6. Step 6: Calculate the equilibrium partial pressure of C2H6C_2H_6: pC2H6=4.0−0.38=3.62p_{C_2H_6} = 4.0 - 0.38 = 3.62 atm.

  7. Step 7: Convert the equilibrium partial pressure to concentration using C=PRTC = \frac{P}{RT}. Here P=3.62P = 3.62 atm, R=0.0821R = 0.0821 atm L mol−1^{-1} K−1^{-1}, and T=899T = 899 K. [C2H6]=3.620.0821×899=0.049[C_2H_6] = \frac{3.62}{0.0821 \times 899} = 0.049 M.

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