Indicate which of these statements about catalysts are true. 1) In an equilibrium process, a catalyst increases the rate of the forward reaction, but leaves the rate of the reverse reaction unchanged. 2) A catalyst is not consumed in the course of a reaction. 3) A catalyst must be carefully chosen to shift the equilibrium toward products.

2) A catalyst is not consumed in the course of a reaction.

Explanation:

A catalyst is a substance which increases the rate of  chemical reaction and remains chemically unchanged at the end of the reaction.  The addition of a catalyst increases the rate of both reversible and irreversible reactions. including reversible reactions involving a final equilibrium state. In reversible reactions, catalysts speed up the time for equilibrium to be attained.

Considering the given statements about catalysts:

1) In an equilibrium process, a catalyst increases the rate of the forward reaction, but leaves the rate of the reverse reaction unchanged.

False: A catalyst increases both the forward and reverse reaction rates equally, thereby allowing the system to reach equilibrium faster.

2) A catalyst is not consumed in the course of a reaction.

True: Catalysts are substances that increase reaction rates without being consumed in the reaction.

3) A catalyst must be carefully chosen to shift the equilibrium toward products.

False: A catalyst does not affect the extent or position equilibrium of a reaction since a catalyst speeds up both the forward and back reactions by  equal amounts , rather it helps a reaction achieve equilibrium 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