An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ At equilibrium, the concentrations are \(\left[\mathrm{H}_{2}\right]=5.0 \mathrm{M},\left[\mathrm{N}_{2}\right]=\) \(8.0 M\), and \(\left[\mathrm{NH}_{3}\right]=4.0 M .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Short Answer

Expert verified
The initial concentrations of nitrogen gas and hydrogen gas were 9 M and 6 M, respectively.

Step by step solution

01

The balanced chemical equation for this reaction is: \( 3 H_2(g) + N_2(g) \rightleftharpoons 2 NH_3(g) \) #Step 2: Write the expression for the equilibrium constant (K) #

The expression for the equilibrium constant (K) can be written as: \( K = \frac{[NH_3]^2}{([H_2]^3 [N_2])} \) #Step 3: Calculate K using the given equilibrium concentrations#
02

We are given the equilibrium concentrations: [ \( H_2 \) ] = 5.0 M [ \( N_2 \) ] = 8.0 M [ \( NH_3 \) ] = 4.0 M Then, calculate K: \( K = \frac{4.0^2}{(5.0^3 \times 8)} \) \( K = \frac{16}{(125 \times 8)} \) \( K = \frac{16}{1000} = 0.016 \) #Step 4: Set up an ICE table (Initial, Change, Equilibrium)#

Let x be the change in the concentration of reactants and products from their initial concentrations to at the equilibrium. [ \( H_2 \) ] [ \( N_2 \) ] [ \( NH_3 \) ] Initial: 5+x 8+x 0-2x Change: -3x -x +2x Equilibrium: 5 8 4 #Step 5: Substitute the concentrations back into the K expression#
03

Substitute the equilibrium concentrations back into the K expression and solve for x: \( K = 0.016 = \frac{(4)^2}{((5+x)^3 \times (8+x))} \) #Step 6: Solve for x and find initial concentrations#

As this equation can be hard to solve algebraically, we will approximate the solution (assuming x is small compared to 5 and 8): \( 0.016 = \frac{16}{(5^3 \times 8)} \) Solve for x: \( x = \frac{16}{(125 \times 8 \times 0.016 )} \) \( x \approx 1 \) Initial concentrations: [ \( H_2 \) ] = 5 + x = 5 + 1 = 6 M [ \( N_2 \) ] = 8 + x = 8 + 1 = 9 M Hence, the initial concentrations of nitrogen gas and hydrogen gas were 9 M and 6 M, respectively.

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Most popular questions from this chapter

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