The diagram shows the potential energy changes for a reaction pathway.

A curved lined graph is shown. The y axis of the graph has the title Potential Energy. The x axis of the graph has the title Reaction Pathway. The graph curve begins at a lower level and ends at a slightly higher level. A vertical line labeled A, starting from the x axis till the beginning of the graph line is shown. A vertical line labeled B, starting from where the line labeled A ended, till the uppermost hump in the graph curve is shown. Another vertical line labeled C is shown from starting from the x axis till the point where the graph curve ends.

Part 1: Describe how you can determine the total change in enthalpy and activation energy from the diagram, and if each is positive or negative.

Part 2: Describe how the curve will look if the reaction was exothermic. Be sure to mention changes in the potential energies of the reactants and products and the sign changes of the enthalpy.

Part 1: You can determine the total change in enthalpy and activation energy from the diagram by the potential energy of the reactants. If the reactants have a high potential energy, then the enthalpy is also high, and if the reactants have a low potential energy, then the enthalpy is low. You can determine if the diagram is positive or negative by knowing if its an endothermic or exothermic reaction.

Part 2: The total change in enthalpy (ΔH) is the difference in potential energy between products and reactants. (P.E. of products < P.E. of reactants).

ΔH = (P.E.) products - (P.E.) reactants.

So, it will be neagtive.

Activation energy is the difference between potential energy of the intermediate and reactants.

It will be positive.

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

the answer is c. nacl