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., $\ce{h2(g) + br2(g) \leftrightarrow 2hbr(g)}$
at equilibrium, concentration values of the compounds are:

compound	$\ce{h2(g)}$	$\ce{br2(g)}$	$\ce{hbr(g)}$

the value of equilibrium constant $k_{eq}$, for the above reaction, is
$\bigcirc$ 2.6
$\bigcirc$ 3.1
$\bigcirc$ 4.3
$\bigcirc$ 5.2
$\bigcirc$ 8.7
question 15 (1 point)
what is the correct equilibrium constant expression for the following reaction?
$\ce{2ncl3(g) \leftrightarrow n2(g) + 3cl2(g)}$
$\bigcirc$ $k_{eq}=\frac{\ce{n2}\ce{cl2}^3}{\ce{ncl3}^2}$
$\bigcirc$ $k_{eq}=\frac{\ce{ncl3}^2}{\ce{n2}\ce{cl2}^3}$
$\bigcirc$ $k_{eq}=\frac{\ce{n2}\ce{cl2}}{\ce{ncl3}}$
$\bigcirc$ $k_{eq}=\frac{\ce{ncl3}}{\ce{n2}\ce{cl2}}$
$\bigcirc$ $k_{eq}=\frac{\ce{n2}^3\ce{cl2}^3}{\ce{ncl3}^2}$

Explanation:

Question 14

Step1: Write the equilibrium constant expression

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

Step2: Substitute the equilibrium concentrations

Substitute \([H_{2}]=0.024\space mol/L\), \([Br_{2}]=0.024\space mol/L\) and \([HBr]=0.05\space mol/L\) into the formula.
\(K_{eq}=\frac{(0.05)^{2}}{0.024\times0.024}\)

Step3: Calculate the value

\(K_{eq}=\frac{0.0025}{0.000576}\approx 4.3\)

For a general reaction \(aA + bB
ightleftharpoons cC + dD\), the equilibrium constant expression is \(K_{eq}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\).
For the reaction \(2NCl_{3}(g)
ightleftharpoons N_{2}(g)+3Cl_{2}(g)\), \(a = 2\), \(c = 1\), \(d=3\).
So \(K_{eq}=\frac{[N_{2}][Cl_{2}]^{3}}{[NCl_{3}]^{2}}\)

Answer:

4.3

Question 15