Co + h_2 + o_2 --> co_2 + h_2_o: delta h= 525 kj

ΔH = -730 kJ

Treat the energy as if it were a product in the equation. You get

CO + H₂ + O₂ → CO₂ + H₂O + 525 kJ

Then you do an ordinary stoichiometry calculation.

Step 1. Convert grams of water to moles of water.

Moles of H₂O = 25 g H₂O × (1 mol H₂O/18.02 g H₂O) = 1.39 mol H₂O

Step 2. Calculate the energy change

ΔH = 1.39 mol water × (-525 kJ/1 mol H₂O) = -730 kJ

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

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