Question

Consider the hypothetical reaction A(g)←→2B(g). A flask is charged with 0.77 atm of pure A, after which it is allowed to reach equilibrium at 0 ∘C. At equilibrium the partial pressure of A is 0.35 atm .A: What is the total pressure in the flask at equilibrium?
B:What is the value of Kp?

Answer 1

Answer:

For A: The total pressure in the flask at equilibrium is 1.19 atm

For B: The value of $K_p$ for the given equation is 2.016

Explanation:

We are given:

Initial partial pressure of A = 0.77 atm

Equilibrium partial pressure of A = 0.35 atm

• For A:

For the given chemical equation:

                  $A(g)\rightleftharpoons 2B(g)$

Initial:         0.77

At eqllm:   0.77-x     2x

Evaluating the value of 'x'

$\Rightarrow (0.77-x)=0.35\\\\x=0.42$

Equilibrium partial pressure of B = 2x = (2 × 0.42) = 0.84 atm

Total pressure in the flask at equilibrium = $p^A_{eq}+p^b_{eq}=[0.35+0.84]atm=1.19atm$

Hence, the total pressure in the flask at equilibrium is 1.19 atm

• For B:

The expression of $K_p$ for given equation follows:

$K_p=\frac{(p_B)^2}{p_A}$

Putting values in above expression, we get:

$K_p=\frac{(0.84)^2}{0.35}\\\\K_p=2.016$

Hence, the value of $K_p$ for the given equation is 2.016