The vapor pressure of ethanol is 400. mmhg at 63.5°c. its molar heat of vaporization is 39.3 kj/mol. what is the vapor pressure of ethanol, in mmhg, at 34.9°c? (r = 8.314 j/k • mol)

The vapor pressure at the new temperature is 352 mmHg.

We can obtain the vapor pressure at different temperatures using the formula;  ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1)).

P1 = initial pressure

P2 = final pressure

ΔHvap = enthalpy of vaporization

R = gas constant

T1 = initial temperature

T2= final temperature

Substituting values;

ln(400/P2) = (39.3 × 10^3 J/mol/ 8.314 J/K.mol) (1/308 - 1/337)

ln(400/P2) = 0.128

(400/P2) = e^0.128

P2 = 400/e^0.128

P2 = 352 mmHg

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-

100 mm Hg

Explanation: -

P 1 =400 mm Hg

T 1 = 63.5 C + 273 = 336.5 K

T 2 = 34.9 C + 273 = 307.9 K

ΔHvap = 39.3 KJ/mol = 39.3 x 10³ J mol⁻¹

R = 8.314 J ⁻¹K mol⁻¹

Now using the Clausius Clapeyron equation

ln (P1 / P2) = ΔHvap / R x (1 / T2 - 1 / T1)

Plugging in the values

ln (400 mm/ P₂) = (39.3 x 10³ J mol⁻¹ / 8.314 J ⁻¹K mol⁻¹) x ( -

= 1.38

P₂ = 100 mm Hg

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

1. they vary with a higher number because boiling is way hotter than melting