Question

consider the following reaction where ( k_{p}=6.25 ) at ( 298 mathrm{~k} ):
( 2 mathrm{no}(mathrm{g})+mathrm{br}_{2}(mathrm{~g})
ightleftharpoons 2 mathrm{nobr}(mathrm{g}) )
if the three gases are mixed in a rigid container at ( 298 mathrm{~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 ( mathrm{nobr}(mathrm{g}) ) is consumed.
2. ( k_{p} ) will increase.
3. a reaction will occur in which ( mathrm{no} ) is produced.
4. ( q ) is greater than ( k ).
5. the reaction is at equilibrium. no further reaction will occur.

Explanation:

Step1: Calculate the reaction quotient \(Q_p\)

The formula for \(Q_p\) for the reaction \(2NO(g)+Br_2(g)
ightleftharpoons 2NOBr(g)\) is \(Q_p=\frac{P_{NOBr}^2}{P_{NO}^2\times P_{Br_2}}\). Given \(P_{NO} = P_{Br_2}=P_{NOBr}=1\ atm\), then \(Q_p=\frac{1^2}{1^2\times1}=1\).

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

Given \(K_p = 6.25\) at \(298\ K\). Since \(Q_p(1)<K_p(6.25)\), the reaction will shift to the right (towards the formation of \(NOBr\)).

Step3: Analyze each statement

• Statement 1: Since the reaction shifts to the right (towards \(NOBr\) formation), \(NOBr\) is not consumed. So, this statement is False.
• Statement 2: \(K_p\) is a function of temperature. Since temperature (\(298\ K\)) is constant, \(K_p\) will not change. So, this statement is False.
• Statement 3: Since the reaction shifts to the right, \(NO\) is consumed (not produced). So, this statement is False.
• Statement 4: As calculated \(Q_p = 1\) and \(K_p=6.25\), so \(Q_p
• Statement 5: Since \(Q_p

eq K_p\), the reaction is not at equilibrium. So, this statement is False.

Answer:

1. F
2. F
3. F
4. F
5. F