Question

A student dissolves of sodium hydroxide in of water in a well-insulated open cup. He that observes the temperature of the water rise from to over the course of minutes. Use this data, and any information you need from the ALEKS Data resource,
NaOH(s) rightarrow Na+(aq) + OH-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1. Is this reaction exothermic, endothermic, or neither.
2. If you said the reaction was exothermic or endothermic, calculate the amount of neat that was released or absorbed by the reaction in this case.
3. Calculate the reaction enthalpy AW per mole of NaOH.

Answer 1

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of the ALEKS Data's resource. However the steps required to answer questions that tests the heat of solution and heat of reaction are all the same.

1. The description provided in the question is an exothermic reaction; this is because exothermic reaction is a reaction that releases/gives off heat into the environment. The heat given-off here is what the student observed as "the temperature of the water rise from to over the course of minutes".

2. To calculate the amount of heat that was released/heat of solution (q) in joules;

q = m × Cg × (T2 - Ti)

where m = mass of water (can also be represented as volume)

Cg = specific heat capacity of water (4.184 joules)

T2 = final temperature of the solution

Ti = initial temperature of the solution

(T2 -Ti) is temperature change of the solution

3. Generally, enthalpy change for exothermic reaction is negative

Hence the formula for enthalpy change/heat of reaction (ΔH);

ΔH = - (q/1000 ÷ number of moles of solute)

Where q is the heat of solution determined in (2) above

number of moles = mass/molar mass of solute

molar mass of NaOH = 23 + 16 + 1 = 40 g/mol (where Na = 23, O = 16 and H = 1)