Which label shows the overall enthalpy change and is the reaction exothermic or endothermic? a, exothermic. b, exothermic. c, endothermic. d, endothermic.

Please, find attached the image that accompanies this question, which is needed to understand and answer it.

Thus, the answer is D, endothermic.

Explanation:

You must assume that the base line (the black horizontal line at the bottom, from which the green and black arrows depart) shows the starting enthalpy with which the reactants of the reaction start.

So, the head of the arrow A shows how high the energy of the reaction has increased from the starting energy(that is the activation energy).

From that point, the energy decreases until the line indicated by the head of the arrow D. So, that is how high is the enthalpy of the products resultant of the chemical reaction compared with the enthalpy of the reactants.

The overall enthalpy change (ΔH rxn) is given by the change in the enthalpy of the products less the enthalpy of the reactants:

That difference, precisely, is what is shown by the arrow labeled D. So, your first conclusion is that the label that shows the overall enthalpy change is the label D.

As for the second question, an endothermic reaction is that that absorbs energy from the surroundings, and the products end with a higher enthalpy than the reactants. That is what the diagram is showing: the products ended at a higher level than the reactants, so energy was absorbed and the reaction is endothermic.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

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