Equilibrium pressure at which 50% of reactant A(g) will be consumed, keeping constant temperature of 300 K inthe given equilibrium is equal

To solve for the equilibrium pressure at which 50% of reactant $A(g)$ will be consumed, we can approach the problem as follows:

1. Break Down the Problem

We want to find the equilibrium pressure $P$ at which the concentration of reactant $A$ is half of its initial concentration ($A_0$).

2. Relevant Concepts

In equilibrium reactions, we can use the equilibrium constant $K_p$ which is expressed in terms of partial pressures. For a general reaction:

$aA \rightleftharpoons bB$

The equilibrium constant $K_p$ is given by:

$K_p = \frac{(P_B)^b}{(P_A)^a}$

Here, $P_A$ and $P_B$ are the equilibrium partial pressures of $A$ and $B$.

3. Analysis and Detail

If we assume that initially $P_A = P_0$ and at equilibrium, 50% of $A$ is consumed, then:

• Initial pressure of $A$: $P_A = P_0$
• Pressure of $A$ at equilibrium: $P_A = \frac{P_0}{2}$
• Pressure of $B$ at equilibrium must be calculated based on the stoichiometry of the reaction.

Assuming $A$ dissociates to form $B$:

• The change in pressure for $A$ is $\frac{P_0}{2}$ (as it decreases by 50%).
• Let’s assume for every 1 mole of $A$ that dissociates, $b$ moles of $B$ are produced. Therefore, the change for $B$ will be $\Delta P_B = \frac{P_0}{2} \times n$ (where $n$ is the stoichiometric coefficient of $B$).

4. Verify and Summarize

We can summarize $K_p$:

$K_p = \frac{(P_B)^b}{(P_A)^a} = \frac{(n \times \frac{P_0}{2})^b}{(\frac{P_0}{2})^a}$

We can solve $K_p$ if we are provided with the value or stoichiometry for the reaction to derive the exact pressure value.

Final Answer

To summarize, the equilibrium pressure at which 50% of reactant $A$ will be consumed requires knowing $K_p$ and the stoichiometric coefficients. The calculations follow from the formulas derived above based on the stoichiometry of the reaction. Please provide the specific reaction or value of $K_p$ for a complete numeric solution.