Question

question 14 (1 point)
hydrogen is the lightest of elements with atomic number one. when hydrogen gas reacts with bromine gas, they form hydrogen bromide gas, i.e., h₂(g) + br₂(g) ↔ 2hbr(g)
at equilibrium, concentration values of the compounds are:

compound	h₂(g)	br₂(g)	hbr(g)
equilibrium concentration	0.024 mol/l	0.024 mol/l	0.05 mol/l

the value of equilibrium constant k_eq, for the above reaction, is
○ 2.6
○ 3.1
○ 4.3
○ 5.2
○ 8.7
question 15 (1 point)
what is the correct equilibrium constant expression for the following reaction?
2ncl₃(g) ↔ n₂(g) + 3cl₂(g)
○ equilibrium constant expression option

Explanation:

Step1: Write the equilibrium constant expression

For the reaction \(H_{2}(g)+Br_{2}(g)
ightleftharpoons 2HBr(g)\), the equilibrium constant expression \(K_{eq}=\frac{[HBr]^{2}}{[H_{2}][Br_{2}]}\)

Step2: Substitute the equilibrium concentrations

Given \([H_{2}]=0.024\space mol/L\), \([Br_{2}]=0.024\space mol/L\) and \([HBr]=0.05\space mol/L\)

Substitute into the formula:

\(K_{eq}=\frac{(0.05)^{2}}{0.024\times0.024}\)

First, calculate \((0.05)^{2}=0.0025\) and \(0.024\times0.024 = 0.000576\)

Then \(K_{eq}=\frac{0.0025}{0.000576}\approx4.3\)

Answer:

4.3