Iron ore is reduced to pure iron by smelting, during which the iron (iii) oxide in the ore reacts with carbon monoxide gas, like this: fe2o3s + 3cog → 2fes + 3co2g suppose an engineer decides to study the rate of this reaction. she prepares four reaction vessels with 158.8g of solid iron (iii) oxide and 33.9g of carbon monoxide gas each. the volume and temperature of each vessel is shown in the table below. arrange the reaction vessels in decreasing order of initial rate of reaction. in other words, select a "1" next to the vessel in which the engineer can reasonably expect the initial rate of reaction to be highest, a "2" next to the vessel in which the initial rate of reaction would be next highest, and so on. rate of reaction a 2.0l 1100.°c b 2.0l 1200.°c c 4.0l 1100.°c d 4.0l 1000.°c

1 b

2 a

3 c

4 d

Explanation:

The reaction rate will be negative. As can seen from the temperature profile, the temperature keeps on increasing. This gives a negative enthalpy of reaction. Thus, = negative.

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