# Question 3dd04

##### 2 Answers
Apr 20, 2017

37.$\text{ }$a) No.
$\text{ }$b) It will shift left. Aka, more reactants will be produced.

38.$\text{ }$${K}_{c} = 2.95 \cdot {10}^{- 2}$

#### Explanation:

37.
First off, you should write the equilibrium expression for the reaction. You should get:
${K}_{c} = 85.0 = \text{[products]"/"[reactants]} = \frac{\left[S {O}_{3}\right] \left[N O\right]}{\left[S {O}_{2}\right] \left[N {O}_{2}\right]}$
Then you just plug in the given concentrations to find out if the reaction is at equilibrium. If the formula equals 85.0 (the equilibrium constant), then the reaction is at equilibrium.

$Q = \frac{\left[0.0400\right] \left[0.0250\right]}{\left[0.00250\right] \left[0.00350\right]}$

Note: $Q$ is what you use when you are unsure if the system is at equilibrium. Otherwise, you use $K$.

$Q = 144.3$

If $Q$ had equaled $K$, then the system is at equilibrium. Since $Q$ is greater than $K$, the reaction has to shift so that the value of $Q$ is decreased. This is accomplished by the formation of more reactants, which will decrease the value of the numerator and increase the value of the denominator.

38.
This question is similar to 37. The first step would be to convert the given moles of $H B r$ into Molarity. So you would get $0.500 \text{M HBr}$.

Then you set up an ICE chart to find the other values at equilibrium.

$\text{ "[HBr]" "[H_2]" } \left[B {r}_{2}\right]$
Initial: $\text{ "0.500M" "0.000M" } 0.000 M$
Change: $\text{ "-2x" "+x" } + x$
Equilibrium: $\text{ "0.500M-2x" "0.000M+x" } 0.0955 M$

Note: $x$ is just a variable. The coefficient before $x$ is based on the reaction coefficients.

Now you just solve for $x$. And from there you find the values of the other gasses at equilibrium.
$\left[H B r\right] = 0.309 M$
$\left[{H}_{2}\right] = \left[B {r}_{2}\right] = 0.0955 M$

Then you just imput those values into the ${K}_{c}$ equation above and you should get:

${K}_{c} = \frac{\left[{H}_{2}\right] \left[B {r}_{2}\right]}{\left[H B r\right]} = \frac{\left[0.0955\right] \left[0.0955\right]}{\left[0.309\right]} = 0.02951537$

Since you only have 3 significant figures, the answer would be:

${K}_{c} = 2.95 \cdot {10}^{- 2}$

39.
This one is almost exactly the same as 38. The only difference is that when you make your ICE chart, the coefficients for $x$ would be different.

~Hope this helps!

Apr 20, 2017

37 (a)The reaction is not at equilibrium; (b) The reaction will have to proceed to the left to reach equilibrium. 38 ${K}_{\textrm{c}} = 0.0955$

#### Explanation:

37. Start by writing the chemical equation with the concentrations beneath the formulas.

$\textcolor{w h i t e}{m m m l} {\text{SO"_2 color(white)(l)+color(white)(m) "NO"_2 color(white)(l)⇌ color(white)(l)"NO" color(white)(l)+ color(white)(l)"SO}}_{3}$
$c : \textcolor{w h i t e}{m} \text{0.002 50"color(white)(m)"0.003 50} \textcolor{w h i t e}{m} 0.0250 \textcolor{w h i t e}{m} 0.0400$

Next. write the equilibrium constant expression (remember, "products over reactants"):

${K}_{\textrm{c}} = \left(\left[{\text{NO"]["SO"_3])/(["SO"_2]["NO}}_{2}\right]\right)$

Insert the concentrations into this expression.

(["NO"]["SO"_3])/(["SO"_2]["NO"_2]) = (0.0250 × 0.0400)/("0.00250 × 0.00350") = 114

(a) We know that ${K}_{\textrm{c}} = 85.0$, so the reaction is not at equilibrium.

(b) The number is bigger than ${K}_{\textrm{c}}$, so we have too many products for equilibrium.

To get to equilibrium, the reaction must get rid of some of the products.

The reaction will shift to the left.

38 For this problem, we can set up an ICE table.

The initial concentration of $\text{HBr" = "2.00 mol"/"4.00 L" = "0.500 mol·L"^"-1}$.

The initial concentrations of ${\text{H}}_{2}$ and ${\text{Br}}_{2}$ are zero, because the reaction hasn't yet started.

We don't know how much hydrogen and bromine will form, but we know it will be some unknown value, $x$.

We also know that, for every mole of hydrogen and bromine formed, 2 mol of $\text{HBr}$ must disappear.

We put all this into the ICE table and get

$\textcolor{w h i t e}{m m m m m m m l} {\text{2HBr" ⇌ "H"_2 + "Br}}_{2}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.500 \textcolor{w h i t e}{m m l} 0 \textcolor{w h i t e}{m m l l} 0$
$\text{C/mol·L"^"-1": color(white)(mm)"-2"xcolor(white)(mml)"+"xcolor(white)(mll)"+} x$
$\text{E/mol·L"^"-1": color(white)(m)"0.500-2} x \textcolor{w h i t e}{m l} x \textcolor{w h i t e}{m m l l} x$

We are told that the equilibrium concentration of $\text{Br"_2 = "0.0955 mol·L"^-1}$.

$x = 0.0955$

The equilibrium concentrations are then

["Br"_2] = "0.0955 mol·L"^"-1"
["H"_2] = x color(white)(l)"mol·L"^"-1" = "0.0955 mol·L"^"1"
["HBr"] = ("0.500 -2"x) color(white)(l)"mol·L"^"-1" = "(0.500 - 2 × 0.0955) mol·L"^"-1" = "0.309 mol·L"^"-1"

Now, we insert these concentrations into the equilibrium constant expression:

K_text(c) = (["H"_2]["Br"_2])/(["HBr"]^2) = (0.0955 × 0.0955)/0.309^2 = 0.0955#