Question

Question 7.10 At 450 K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.

2SO2(g) + O2(g) <---> 2SO3(g)

What is Kc at this temperature?

Class XI Equilibrium Page 225

Answer 1

2SO2(g) + O2(g) <=> 2SO3(g)

we know,
$K_p =K_c(RT)^{\triangle{n}_g}$
where, ∆ng = number of moles of gaseous products - number of moles of gaseous reactants = 2 - (1 + 2) = -1

Kp = 2 × 10^10 /bar
R = 0.0831 L.bar/K/mol
T = 450 K

Kc = Kp/(RT)^(-1) [ because , ∆ng = -1 ]
= Kp(RT)
= 2 × 10^10 × 0.0831 × 450
= 7.479 × 10¹¹ L/mol

Answer 2

Answer:

7.48×10¹¹ L/mol

Explanation:

2SO2(g) + O2(g) <=> 2SO3(g)

Here, ∆ng = number of moles of gaseous products - number of moles of gaseous reactants = 2 - (1 + 2) = -1

Kp = 2 × 10^10 /bar

R = 0.0831 L.bar/K/mol

T = 450 K

Kc = Kp/(RT)^(-1) [ because , ∆ng = -1 ]

= Kp(RT)

= 2 × 10^10 × 0.0831 × 450

= 7.479 × 10¹¹ L/mol