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3 years ago

The following system is at equilibrium:

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

A. Rightward shift.

Adding more X will increase the reactants which will inturn produce more product. i.e rightward.

B. Leftward shift.

Removing X means reducing the reactant and this will reduce the reaction, thereby making the products to reverse back to reactants. i.e leftward.

C. Leftward shift.

Increasing volume means decreasing the pressure. This will make the reaction to proceed to where there is greater volume. i.e leftward.

D. Rightward shift

Decreasing the volume means increasing the pressure. The reaction will proceed to where there is a decrease in volume i.e rightward.

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Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

When one mole of sodium chloride dissociates in water, it produces: 1 mole of ions ½ mole of sodium ions 2 moles of ions ½ mole

adelina 88 [10]

Dissociation of NaCl in water is given as below,

NaCl ₍s₎ → Na⁺ ₍aq₎ + Cl⁻ ₍aq₎

According to this balanced equation the moles on RHS and LHS are as,

Moles on LHS,
NaCl ₍s₎ = 1 Mole

Moles on RHS,
Na⁺ ₍aq₎ = 1 Mole

Cl⁻ ₍aq₎ = 1 Mole
Result:
When one mole of sodium chloride dissociates in water, it produces 2 Moles of Ions.

7 0

3 years ago

Read 2 more answers

Boyles Law-

My name is Ann [436]

Answer:

Boyles Law-

Variables that change: volume and pressure

Variables that remain constant: Temperature

Charles Law-

Variables that change: volume and temperature

Variables that remain constant: pressure