Amixture of cs2(g) and excess o2(g) is placed in a 10 l reaction vessel at 100.0 ∘c and a pressure of 3.15 atm . a spark causes the cs2 to ignite, burning it completely, according to the equation: cs2(g)+3o2(g)→co2(g)+2so2(g) after reaction, the temperature returns to 100.0 ∘c, and the mixture of product gases (co2, so2, and unreacted o2) is found to have a pressure of 2.50 atm . what is the partial pressure of each gas in the product mixture? enter your answers numerically separated by commas.

0.65, 1.30, 0.55

Explanation:

By the Raoult's Law, the partial pressure of a gas in a gas mixture is its molar fraction multiplied by the total pressure of the mixture. Let's found out the total number of moles of the mixture using the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (100°C + 273 = 373 K).

3.15*10 = n*0.082*373

30.586n = 31.5

n = 1.03 moles of the reactants

Let's call the total number of moles of CS₂ as x and the number of moles of O₂ is 1.03 - x, thus by the stoichiometry of the reaction, 1:3:1:2, the number of moles of O₂ that reacts is 3x, and the remaining number of moles of it, and the number of moles of each product is:

nO₂ = (1.03 - x) - 3x

nCO₂ = x

nSO₂ = 2x

After the reaction happened, the number of moles is:

2.50*10 = n*0.082*373

30.586n = 25.0

n = 0.8174 mol

Thus,

nO₂ + nCO₂ + nSO₂ = 0.8174

(1.03 - x) -3x +x + 2x = 0.8174

1.03 - x = 0.8174

x = 0.2126 mol

nO₂ = (1.03 - 0.2126) - 3*0.2126 = 0.1796 mol

nCO₂ = 0.2126 mol

nSO₂ = 2*0.2126 = 0.4252 mol

Using Roult's Law, the partial pressures will be:

pCO₂ = (nCO₂/n)*2.50 = (0.2126/0.8174)*2.50 = 0.65 atm

pSO₂ = (nSO₂/n)*2.50 = (0.4252/0.8174)*2.50 = 1.30 atm

pO₂ = (nO₂/n)*2.50 = (0.1796/0.8174)* 2.50 = 0.55 atm

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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