Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. predict the effect (shift right, shift left, or no effect) of the indicated volume change. co(g) + h2o(g) < => co2(g) + h2(g) (volume is decreased) pcl3(g) + cl2(g) < => pcl5(g) (volume is increased) caco3(s)< => cao(s) + co2(g) (volume is increased)

CO(g) + H₂O(g) <=> CO₂(g) + H₂(g), (volume is decreased) .. No effect.

PCl₃(g) + Cl₂(g) <=> PCl₅(g) , (volume is increased)               .. Shift left.

CaCO₃(s) <=> CaO(s) + CO₂(g) , (volume is increased)         .. Shift right.

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

CO(g) + H₂O(g) <=> CO₂(g) + H₂(g)            (volume is decreased)

So, decreasing the volume will have no effect on the reaction.

PCl₃(g) + Cl₂(g) <=> PCl₅(g) , (volume is increased)

so, increasing the volume will shift the reaction left.

CaCO₃(s) <=> CaO(s) + CO₂(g) , (volume is increased)

so, increasing the volume will shift the reaction right.

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