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PIT_PIT [208]
3 years ago
9

Which equation, in which k is a constant, represents an inverse proportion between the variables? V = T/k F = –kx2 PV = k y = kx

– 8
Chemistry
1 answer:
Alona [7]3 years ago
5 0

The inverse proportion is PV = k

We can write the equation as <em>P</em> = k/<em>V</em> or <em>P</em> ∝ 1/<em>V</em>. Thus, <em>P</em> and <em>V</em> are <em>inversely related</em>.

We can rewrite <em>V</em>= T/k as <em>V</em> = k<em>T</em> or <em>V</em> ∝ <em>T</em>. Thus,<em> V</em> is <em>directly proportiona</em>l to <em>T</em> .

<em>F</em> = -k<em>x</em>² is <em>neither a direct nor an invers</em>e relation.

<em>y</em> = k<em>x</em> – 8 is a linear relation, but it is <em>not a direct proportion</em> because of the “-8” term.

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What is the molar mass of a gas if a flask with a volume of 3. 16 l contains 9. 33 grams of the gas at 32. 0°c and 1. 00 atm?
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The molar mass of a gas if a flask with a volume of 3. 16 L contains 9. 33 grams of the gas at 32. 0°C and 1. 00 atm is  1.17g/mol

Calculation ,

In this question we have to fist find the number of moles of gas by using ideal gas equation and from the help of number of moles we can determine molar mass.

According to ideal gas equation which is also known as ideal law ,

PV = nRT                ...( i )

where P is the pressure of the gas = 1 atm

V is the volume of the gas in the flask with volume =  3. 16 L

R is the universal gas constant = 0.082 atm L/K mol

T is the temperature = 32. 0°C = 32 + 273 = 305 K

n is the number of moles = ?

Putting the value of Pressure P , volume V , temperature T , number of moles n and universal gas constant R in the equation (i) we get ,

1 atm ×3. 16 L = n× 0.082 atm L/K mol ×305 K

n = 1 atm ×3. 16 L / 0.082 atm L/K mol × 305 K = 0.126 mole

number of mole of a gas  = 0.126 mole = given mass/ molar mass

molar mass  = number of moles × Given mass =  0.126 × 9. 33 = 1.17g/mol

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
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