All categories

3 years ago

Consider the following equilibrium:

Chemistry

1 answer:

Ainat [17]3 years ago

7 0

<u>Answer: </u>The equation which is wrong is $K_p=K_c(RT)^{-5}$

<u>Explanation:</u>

For the given reaction:

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

The expression for $K_c\text{ and }K_p$ is given by:

$K_c=\frac{1}{[O_2]^3}$

$K_p=\frac{1}{[O_2]^3}$

The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.

Relationship between $K_p\text{ and }K_c$ is given by the expression:

$K_p=K_c\times (RT)^{\Delta n_g}$

where,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants

R = gas constant

T= temperature

For the above reaction,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3

Hence, the expression for $K_p$ is:

$K_p=K_c\times (RT)^{-3}$

Therefore, the equation which is wrong is $K_p=K_c(RT)^{-5}$

You might be interested in

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

3 years ago

the amount of solar energy reaching 1 m² of earths surface each second is 1.366 x 10^3 joules. if this energy was converted into

Advocard [28]

I think is 6.588579795x10^13 Kg because the equation is E=mc^2 and E is your Joules and c^2= 9x10^16 so m=(c^2)/E

4 0

3 years ago

What does the last line of the poem suggest

Klio2033 [76]

Answer:

that the poem needs to be finished heh pog

Explanation:

7 0

3 years ago

Read 2 more answers

Sunny_sXe [5.5K]

So a camera can zoom in and out and the human eye has peripheral vision

4 0

3 years ago

Read 2 more answers

Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

$K_{sp}$ = $5.02\times 10^{-6}$

First we have to calculate the solubility of $OH^-$ ion.

The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Let the solubility will be, 's'.

The concentration of $Ca^{2+}$ ion = s

The concentration of $OH^-$ ion = 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][OH^-]^2$

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

$pOH=1.67$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33$

Therefore, the pH of a saturated solution is, 12.33

7 0

3 years ago