For NH 4 HS (s) leftharpoons NH 3(g)+ H 2 S (g), the observed pressure for the reaction mixture in equilibrium is 1.12 atm at 106° C. What is the value of Kp for the reaction?

Question

For NH 4 HS (s) leftharpoons NH 3(g)+ H 2 S (g), the observed pressure for the reaction mixture in equilibrium is 1.12 atm at 106° C. What is the value of Kp for the reaction? (A)  0.56 atm2  (B)  0.3136 atm2  (C)  1.25 atm2  (D)  1.12 atm2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/21e1869d0081a88c2ddaca920faac12a-.png" /> p texttotal=1.12 atm Kp=pNH3× pH2 s pNH3=p texttotal × xNH3 =1.12×(x/2x)=0.56 pH2=p texttotal× xH2 s =1.12×(x/2x)=0.56 ∴ Kp=0.56×0.56=0.3136 atm2

Q. For $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ , the observed pressure for the reaction mixture in equilibrium is $1.12$ atm at $10 6^{\circ} C$ . What is the value of $K_{p}$ for the reaction?

$0.56 a t m^{2}$

$0.3136 a t m^{2}$

$1.25 a t m^{2}$

$1.12 a t m^{2}$

$p_{total} = 1.12 a t m$
$K_{p} = p_{N H_{3}} \times p_{H_{2} s}$
$p_{N H_{3}} = p_{total} \times x_{N H_{3}}$
$= 1.12 \times \frac{x}{2 x} = 0.56$
$p_{H_{2}} = p_{total} \times x_{H_{2} s}$
$= 1.12 \times \frac{x}{2 x} = 0.56$
$∴ K_{p} = 0.56 \times 0.56 = 0.3136 a t m^{2}$

Q. For NH4​HS(s)⇌NH3​(g)+H2​S(g), the observed pressure for the reaction mixture in equilibrium is 1.12 atm at 106∘C. What is the value of Kp​ for the reaction?

0.56atm2

0.3136atm2

1.25atm2

1.12atm2