Explanation Given: T=700 K; K n =1.60 10 -3 ;V=1.55 L;P H2O =145 torr Formula used: K p = [ PH2 ] [ PCO ] [ PH2O ] Calculation: We can begin by writing out the reaction that is taking place and then prepare an ICE table to solve for the equilibrium partial pressures: C(s)+H 2 O(g)⇌CO (g) +H 2(g) H 2 O H 2 CO I 145 0 0 C - x +x + x E 145– x x x We can insert these values into the equilibrium expression to solve for x . Since, we have a large initial pressure and a small value of K p , we can use the x is small approximation: K p = [ PH2 ] [ PCO ] [ PH2O ] 1.60 × 10 − 3 = ( x ) ( x ) ( 145 − x ) 1.60 × 10 − 3 = x 2 145 x = 0.232 x=0.48 torr The partial pressure of H 2 at equilibrium is 0.48 torr. We are asked for the percent of H 2 gas by mass, therefore, we need to find the masses of each gas in the system. First we will calculate the mass of H 2 ...

Pearson eText for Chemistry: structures and Properties -- Instant Access (Pearson+)

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Author: Nivaldo Tro

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Chapter 15, Problem 92E

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To determine: Find the percent by mass of hydrogen gas present at equilibrium.

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Chapter 15, Problem 91E

Chapter 15, Problem 93E

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Pearson eText for Chemistry: structures and Properties -- Instant Access (Pearson+)

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Find the percent by mass of hydrogen gas present at equilibrium.

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To determine: Find the percent by mass of hydrogen gas present at equilibrium.

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