0, (s)+ 120() 12co,(g)+ 11 H,0 (g) -5155.7 k c. (3 ports) The reaction in part b

Question

0, (s)+ 120() 12co,(g)+ 11 H,0 (g) -5155.7 k c. (3 ports) The reaction in part b occurs in a cannister, originally at 20.00 °C What is final temperature in the cannister after the reaction (assume that by now, it contains only 115.8g water and 308.6 g carbon dioxide)' Hint: first, solve for change in temperature.co c, M,o=1.864 g°C d. (1 points) It, as in part c115.8 g water vapor and 308.6 g carbon dioxide are formed, what is the partial pressure of carbon dloxide? Assume the cannister has a total pressure of 7290 atm In the film, the cannister explodes, releasing the contents into the vacuum of space. Imagine, instead, that our cannister springs a leak e. (1 point) Which would escape more quickly: water vapor / carbon dioxide? Circle the correct answer and explain your answer in a couple of sentences vi. vii. (2 points) Prove your answer in part calculate the effusion rate for water vapor, if carbon dicxide's effusion rate is 1.25 x 10 m/s.

Explanation / Answer

c. The reaction involved in the question is:

C12H22O11 (s) + 12 O2 (l) ----------> 12 CO2 (g) + 11 H2O (g) delta Hrxn = -5155.7 KJ

Initial temperature = 20.0 deg C

Mass of CO2 produced = 308.6 g

Mass of H2O ?produced = 115.8 g

delta Hrxn = -5155.7 KJ => heat released when 1 mol of C12H22O11 is burnt or heat absorbed when 12 moles CO2 is produced

So we should first calculate the amount of heat aborbed by 308.6 g CO2

moles of CO2 = 308.6 g CO2 * (1 mole CO2/ 44.0 g /mol) = 7.014 moles CO2

Heat produced by CO2 = 7.014 moles CO2 * (delta H/ 12 moles CO2)

= 7.01 moles CO2 * ( -5155.7 KJ/ 12 moles CO2) = -3013.4 KJ

mass of CO2 * specific heat of CO2 * (final temp - initial temp) = heat produced

308.6 g * 0.846 J.g. deg C *  (final temp - 20) deg c = -3013.4 KJ

(final temp - 20) = -3011 KJ/308.6*0.846 = -11.5 deg C

final temp = -11.5 + 20 = 8.5 deg C

d. If Mass 308.6 g of CO2 and 115.8 g of H2O ?are produced and total pressure is 7290 atm. What is the partial pressure of CO2?

moles H2O ?= 115.8/18.00 = 6.433 moles

moles CO2 =308.6/44.00 = 7.014 moles

partial pressure of CO2 = mole fraction of CO2 * total pressure

= moles of CO2 /(moles ofCO2 + molesH2O ?) * 7290 = 7.014/ (7.014 + 6.433) * 7290 = 3802 atm

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