All categories

2 years ago

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

Chemistry

2 answers:

gtnhenbr [62]2 years ago

8 0

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 initial partial pressure

-x -x +x +x react

105.1-x 7.96-x x x equilibrium

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

anzhelika [568]2 years ago

6 0

Answer:

0.422 torr

Explanation:

First you want to set up a table with three rows. Initial partial pressure, change in partial pressure, and equilibrium partial pressure. Make four columns, CH4, I2, CH3I,and HI. You are given the initial pressures of 105.1 torr and 7.96 torr for methane and I2. Put these in the respective row and column. In terms of change, CH4 would be -x due to the coefficient of 1, I2 would be -x as well. CH3I and HI would both be +x in terms of change. Initial+ change = equilibrium. So the equilibrium of methane is 105.1 torr-x and I2 is 7.96-x. Since the initial pressures of CH3I and HI are not given, they can be assumed to be zero, and therefore have equilibrium of +x. Kp = [HI][CH3I]/[CH4][I2]. Therefore, Kp= x^2/ (105.1-x)(7.96-x). Plug in the value of Kp and solve for x. The value of x should be roughly .42217 M. Now, since CH4 is 105.1-x, then 105.1-.42217= 104.7 torr. Following suit, I2= 7.96-.x and therefore 7.96-.42217= 7.54 torr. Since the equilibrium of CH3I and HI is just +x, the value of the equilibrium will just be .42217.

So:

CH4= 104.7 torr

I2= 7.54 torr

CH3I and HI= .422 torr

You might be interested in

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

1 year ago

Some fruits and vegetables are preserved by pickling them. Nandini got confused

stepan [7]

Answer:

Explanation:

3 0

1 year ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.