At 700 K, the equilibrium constant Kp for the reaction 2 SO 3(g) leftharpoons 2 SO 2(g)+ O 2(g) is 1.80 × 10-3 k Pa. What is the numerical value of Kc in moles per litre for this reaction at the same temperature?

Question

At 700 K, the equilibrium constant Kp for the reaction 2 SO 3(g) leftharpoons 2 SO 2(g)+ O 2(g) is 1.80 × 10-3 k Pa. What is the numerical value of Kc in moles per litre for this reaction at the same temperature? (A)  3.13× 10-7  (B)  3.13× 107  (C)  3.13× 10-10 (D) None of these

Tardigrade Admin · Accepted Answer

Here np=3 mol, nr=2 mol ∴ Δ ng =np-nr=3-2=1 . mol Kp =1.80 × 10-3 k Pa =1.80 Pa =(1.80/105) bar =1.80 × 10-5 bar ∵ Kp=Kc(R T)Δ ng Kc=(Kp/R T) =(1.80 × 10-5 bar /0.0821 L bar K -1 mol -1 × 700 K ) =3.13 × 10-7 mol L -1

Q. At $700 K$ , the equilibrium constant $K_{p}$ for the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
is $1.80 \times 1 0^{- 3} k P a$ . What is the numerical value of $K_{c}$ in moles per litre for this reaction at the same temperature?

$3.13 \times 1 0^{- 7}$

$3.13 \times 1 0^{7}$

$3.13 \times 1 0^{- 10}$

None of these

Q. At 700K, the equilibrium constant Kp​ for the reaction 2SO3​(g)⇌2SO2​(g)+O2​(g) is 1.80×10−3kPa. What is the numerical value of Kc​ in moles per litre for this reaction at the same temperature?

3.13×10−7

3.13×107

3.13×10−10

None of these

Here np​=3 mol, nr​=2 mol ∴Δng​=np​−nr​=3−2=1. mol Kp​=1.80×10−3kPa =1.80Pa=1051.80​ bar =1.80×10−5 bar ∵Kp​=Kc​(RT)Δng​ Kc​=RTKp​​ =0.0821LbarK−1mol−1×700K1.80×10−5bar​ =3.13×10−7molL−1

Q. At $700 K$ , the equilibrium constant $K_{p}$ for the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
is $1.80 \times 1 0^{- 3} k P a$ . What is the numerical value of $K_{c}$ in moles per litre for this reaction at the same temperature?

$3.13 \times 1 0^{- 7}$

$3.13 \times 1 0^{7}$

$3.13 \times 1 0^{- 10}$