**Problem 10: Equilibrium in a Chemical Reaction**A flask is charged with 2.50 atm of A(g) and 7.50 atm of B(g) at 30°C and the following equilibrium is achieved: \[ 2A(g) \rightleftharpoons 3B(g) \]After equilibrium is reached, the partial pressure of A is 0.375 atm.**Questions:**a. What is the equilibrium partial pressure of B?b. Calculate the $ K_p $ at 30°C.

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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**Problem 10: Equilibrium in a Chemical Reaction**

A flask is charged with 2.50 atm of A(g) and 7.50 atm of B(g) at 30°C and the following equilibrium is achieved:

\[ 2A(g) \rightleftharpoons 3B(g) \]

After equilibrium is reached, the partial pressure of A is 0.375 atm.

**Questions:**

a. What is the equilibrium partial pressure of B?

b. Calculate the $ K_p $ at 30°C.

Expert Solution

Step 1

Given:

Initial partial pressure of A = 2.50 atm

Initial partial pressure of B = 7.50 atm

Equilibrium partial pressure of A =0.375 atm.

Step 2

(a) The equilibrium reaction is,

$2 A (g) \to 3 B (g)$

The ICE table for pressure is,

	$2 A (g) \to$	$3 B (g)$
Initial	2.50 atm	7.50 atm
Change	-2x	+3x
Equilibrium	(2.50-2x) = 0.375 atm	(7.50+3x)

From the ICE table calculate value for variable x as,

$\begin{array}{rcl} 2.50 a t m - 2 x & = & 0.375 a t m \\ 2.50 a t m - 0.375 a t m & = & 2 x \\ 2.125 a t m & = & 2 x \\ x & = & 1.0625 a t m \end{array}$

By using value for x, calculate equilibrium partial pressure of gas B as shown below,

$7.50 + (3 \times 1.0625) = 10.69 a t m$

Thus, the partial pressure of B at equilibrium is 10.7 atm.

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