Question

Consider the following reaction:

2SO2(g) + O2(g) 2SO3(g)
If the value of Kc for this reaction is 13 at 900 K, calculate thevalue of K at 900 K.
a) 0.18 b) 960 c) 13 d) 0.0024 e) 0.077

2. The equilibrium constant for the reaction
2BrCl(g) Br2(g) + Cl2(g)
is 32 at 500 K. A mixture of BrCl, Br2 and Cl2, each at 0.050 M,was introduced into a container
at 500 K. Which of the following is true?
a) The reaction proceeds until all BrCl(g) is used up.
b) The system is at equilibrium and therefore no net changeoccurs.
c) Br2(g) is used up.
d) The reaction proceeds to the right, forming more products.
e) The reaction proceeds to the left, forming more BrCl(g).

Answer 1

1. a) 0,18

2. e) The reaction proceeds to the left, forming more BrCl(g).

Step-by-step explanation:

1. In a gas reaction as:

2SO₂(g) +O₂(g) ⇄ 2SO₃(g)

it is possible to convert kp to kc using:

kp = kc (RT)^Δn

Where kp is gas equilibrium constant, kc is equilibrium constant (13), R is gas constant (0,082atmL/molK), T is temperature (900K), and Δn is number of moles of gas products - number of moles of gas reactants (That is 2 - (2+1) = -1). Replacing:

kp = 13×(0,082atmL/molK×900K)^-1

kp = 0,18

2. Based on Le Chatelier's principle, the change in temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system. For the reaction:

2 BrCl(g) ⇄ Br₂(g) + Cl₂(g).

The addition of 0,050M of each compound cause the reaction proceeds to the left, forming more BrCl(g) because based on the reaction, you need two moles of BrCl per mole of Br₂(g) and Cl₂(g) to keep the system in the same. But you are adding the same proportion of moles of each compound.

I hope it helps!