Question

consider the following reaction where ( k_{p}=1.80\times10^{-2} ) at 698 k.

( 2hi(g)
ightleftharpoons h_{2}(g)+i_{2}(g) )

if the three gases are mixed in a rigid container at 698 k so that the partial pressure of each gas is initially one atm, what will happen?

indicate true (t) or false (f) for each of the following:

1. a reaction will occur in which ( hi(g) ) is produced.

1. ( k_{p} ) will decrease.

1. a reaction will occur in which ( h_{2} ) is consumed.

1. ( q_{p} ) is greater than ( k_{p} ).

1. the reaction is at equilibrium. no further reaction will occur.

Explanation:

Step1: Calculate \(Q_p\)

The reaction quotient \(Q_p\) for the reaction \(2HI(g)
ightleftharpoons H_2(g)+I_2(g)\) is given by \(Q_p=\frac{P_{H_2}\times P_{I_2}}{P_{HI}^2}\).
Since \(P_{H_2} = P_{I_2}=P_{HI} = 1\ atm\), then \(Q_p=\frac{1\times1}{1^2}=1\).

Step2: Compare \(Q_p\) and \(K_p\)

Given \(K_p = 1.80\times10^{-2}\).
Since \(Q_p(1)>K_p(1.80\times 10^{-2})\), the reaction will shift to the left.

Step3: Analyze each statement

1. For statement 1:

Since the reaction shifts to the left (reverse reaction), \(HI(g)\) is produced. So, statement 1 is \(T\).

1. For statement 2:

\(K_p\) is a constant at a given temperature. It does not change with the concentrations or partial pressures of reactants and products. So, statement 2 is \(F\).

1. For statement 3:

Since the reaction shifts to the left, \(H_2\) (a product in the forward reaction) is consumed. So, statement 3 is \(T\).

1. For statement 4:

As calculated \(Q_p = 1\) and \(K_p=1.80\times 10^{-2}\), so \(Q_p>K_p\). Statement 4 is \(T\).

1. For statement 5:

Since \(Q_p
eq K_p\), the reaction is not at equilibrium. Statement 5 is \(F\).

Answer:

1. \(T\)
2. \(F\)
3. \(T\)
4. \(T\)
5. \(F\)