Question

1. The decomposition reaction of carbon disulfide, CS2, to carbon monosulfide, CS, and sulfur is first order with k = 2.8x10-7 s -1 at 1000˚C. CS2....> CS + S

a) What is the half life of this reaction at 1000˚C?

b) How many days would pass before a 2.00 g sample of CS2 had decomposed to the extent that 0.75 g of CS2 remains?

c) Refer to part b). How many grams of CS would be present after this length of time?

d) How much of a 2.00 g sample of CS2 would remain after 45.0 days?

2- The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k = 0.80 M-1 ∙s -1 . 2 NOBr(g)......>2 NO(g) + Br2(g)

a) What is the half-life of this reaction if the initial concentration of NOBr is 0.68 M?

b) How much time would pass before a 2.00 g sample of NOBr held in a 0.255 L flask had decomposed to the extent that 0.65 g of NOBr remains?

c) Refer to part b). How many grams of Br2 would be present after this length of time?

d) How much of a 2.00 g sample of NOBr would remain after thirteen seconds?

3- a) A student performed a calorimetry experiment. He combined 50.0 mL of water at 85˚C and 50.0 mL of water at 22˚C in a coffee cup calorimeter. The final temperature of the water was 45˚C. What was the heat capacity of the calorimeter?

b) The same calorimeter was used to find the specific heat of a metal. He heated 95.22 grams of the metal in a boiling water bath (99˚C). After adding this hot metal to 100.0 mL of 21˚C water in the calorimeter, the final temperature reached 29˚C. What is the metal’s specific heat?

Answer 1

1. The decomposition reaction of carbon disulfide, CS2, to carbon monosulfide, CS, and sulfur is first order with k = 2.8x10-7 s -1 at 1000˚C. CS2....> CS + S

a) What is the half life of this reaction at 1000˚C?

if this is 1st order, then we can get Half life, HL, as follows:

HL = ln(2) / k

HL = ln(2) / (2.8*10^-7)

HL = 2475525.64 seconds --> 2475525.64/3600 h = 687.64 h = 687.64/24 = 28.65 days

b) How many days would pass before a 2.00 g sample of CS2 had decomposed to the extent that 0.75 g of CS2 remains?

For first order

dC/dt = k*C^1

dC/dt = k*C

When developed:

dC/C = k*dt

ln(C) = ln(C0) - kt

ln(0.75) = ln(2) - (2.8*10^-7)*t

t = ( ln(0.75) - ln(2) ) / - (2.8*10^-7) = 3502961.6179 seconds

t = 3502961.6179 s * 1day / 86400 s

t = 40.5 days

c) Refer to part b). How many grams of CS would be present after this length of time?

Mass of CS --> from mol stoihciometry

mol of CS2 reacted = mass of CS2 / MW = (2-0.75)/(76.1407 ) =0.01641 mol reacted

ratio is 1:1

mol of CS formed = .01641 mol

mass = mol*MW = (0.01641*44) = 0.72204 g

d) How much of a 2.00 g sample of CS2 would remain after 45.0 days?

after t = 45 days

t = 45 *86400 s = 3.888*10^6 s

ln(C) = ln(C0) - kt

lnC = ln(2) - (2.8*10^-7)(3.888*10^6)

C = exp(-0.39549) = 0.673350 g left