A potential energy diagram is shown. A graph is plotted with Potential Energy in KJ on the y axis and Reaction Pathway on the x axis. The graph starts at 30 on the y axis and it is initially horizontal, labeled Reactants, and then rises up as a smooth curve and reaches its peak when y is 65 kilojoules. The curve comes down until y is 55 kilojoules where it becomes horizontal again, and it is labeled Products. What is the activation energy of this reaction? (5 points) 10 kJ 25 kJ 30 kJ 35 kJ

35 KJ.

Explanation:

The activation energy is the minimum energy that must be overcome for a reaction to take place.

In the diagram given above, the activation energy lies between the energy of the reactants and that at the peak.

Thus we can calculate the activation energy as follow:

Energy of reactants = 30 KJ

Energy at the peak = 65 KJ

Activation energy =..?

Activation energy = Energy at the peak – Energy of reactants

Activation energy = 65 – 30

Activation energy = 35 KJ

Therefore, the activation energy of th reaction is 35 KJ

the answer is   c.) 1575 j.

explanation:

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