A solution consists of 0.750 mole of pentane, C_3 H_12, and 0.150 cyclopentane C

Question

A solution consists of 0.750 mole of pentane, C_3 H_12, and 0.150 cyclopentane C_3 H_15. The partial pressure of pentane in the vapor that is in equilibrium with the solution is torr. the vapor pressures of the pure liquids at 25 degree C are 451 torr for pentane and 323 torr for cyclopentane. Assume that the solution is an ideal solution. a) 119 b) 267 c)376 d) 338 e) 75.2 Given the equation for the following reaction occurring at 225 degree C BaO_2 (s) + 4 HCI (g) BaCI_2 (s) + 2 H_2 O(g) + CI_2 (g) The K_c expression for the equilibrium is: a) [BaO_2][HCI]^4/[BaCI_2]{H_2 O]^2[CI_2] b) [H_2 O]^2 [CI_2]/[HCi]^4 c) [Ci_2]/[HCI]^4 d) [H_2 O]^2 [CI_2]/[BaO_2][HCI]^4 e) [H_2 O]^2 + [CI_2]/[HCI]^4 A.3.00 L flask initially contained 0.200 mol CO_2, 0.500 mol CO, and 0.800 mol O_2 at 2000 K. Given that the equilibrium constant (K_c) for the reaction: 2 CO_2 2 CO + O_2 is 0.738 at 2000 K. Which of the following statements is true? The value of "Q" is and the. a) 0.333, concentration of CO_2 will decrease as the system approaches equilibrium b) 1.67, concentration of CO_2 will increase as the system approaches equilibrium c) 0.600 concentration of CO_2 will increase as the system approaches equilibrium. d) 5.00, concentration of CO_2 will decrease as the system approaches equilibrium e) 5.00, concentration of CO_2 will increase as the system approaches equilibrium

Explanation / Answer

Q6

0.75 mol of pentane, 0.15 mol of cyclopentane

total mol = 0.75+0.15 = 0.90

mol fraciton of pentane = 0.75/0.90 = 0.83333

mol fraction of cyclopentane = 0.15/0.90 = 0.16666

so...

The partial pressure of pentane in vapor pressure...

if P = 321 torr for cycloe

if P = 451 torr for pentane

then

Ptotal = 321*0.15 + 0.85*451 = 431.5 torr

so..

for pentane = 0.85*451 = 383.35 torr

nearest answer is D

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