Assume that 100.0 mL of 0.200 M CsOH and 50.0 mL of 0.400 M HCl a...

Question

Assume that 100.0 mL of 0.200 M CsOH and 50.0 mL of 0.400 M HCl are mixed in a calorimeter. The solutions start out at 22.50 °C, and the final temperature after reaction is 24.28 °C. The densities of the solutions are all 1.00 g/mL, and the specific heat of the mixture is 4.2 J/(g•°C). What is the enthalpy change for the neutralization reaction of 1.00 mol of CsOH in kJ?

Accepted Answer

Hey everyone in this example we have a 20 g sample of butane that is combusted. And the heat from the reaction is used to raise the temperature of five liters of water from 25°C initially to 72.37°C the end. And so we need to calculate the entropy of our reaction for the combustion of butane. So our first step is to recognize that because we have a combustion reaction and the heat being used to heat up this water here, we're technically going to be calculating the amount of heat that water absorbs. And we know that it's absorbing heat because it goes from a very low initial temperature to a much higher final temperature. So what we want to do is recall our formula for heat which is Q. For heat equal to mass times the heat capacity of our substance times the difference in temperature, delta T. Where we recall that delta T. Is the final minus the initial temperature. So what we want to do is get the mass of water because they were only given water as a volume in leaders. So we want to convert From five L into g. And so we want to recall that the density of water Is one g per middle leader. And so we're going to use that as a conversion factor where we will begin with our 5l of water. So this is to find the mass of water and grams. And we're going to go from leaders into middle leaders. So we want to recall that our prefix Milli tells us that we have 10 to the negative. Three leaders in one middle leader. Next we want to go from middle leaders into graham's using that density of water. So we know that we have for one millimeter of water one g of water. And so we can cancel out our units. Middle leaders and leaders leaving us with grams. And we get a total of 5000 g of water for our massive water. So now that we have the massive water in the proper units, let's go ahead and calculate the heat that water absorbs from our combustion reaction of butane. So we have Q. is equal to 5000 g Multiplied by the specific heat of water, which we recall is 4.184 jules divided by grams times degrees Celsius. And then this is multiplied by Delta T. Where we have the final temperature given as 72.37°C minus the initial temperature given as 25°C. And so what we're going to recognize is that we're going to cancel out our units of gramps as well as the degrees Celsius units leaving us with jewels. And so what we'll have is that Q. Is equal to 990, 980. Sorry about that jules. And so we want to make sure that this unit is actually in kilo joules. So we're going to recall our conversion factor to go from jules tequila jewels. where the prefix kilo tells us that for one jewel we have 10 to the negative third power kilo jewels. And this gives us a total After we cancel out the jewels were left with kilo jewels, we have a total of 991 kg jewels as our heat that is absorbed by our water. And so our next step is to find moles of our butane. And in order to do so, we're going to use that mask that were given, which is 20 g of butane. And then we're going to refer to our periodic tables to recall the molar mass for butane. So grams per mole. And so looking at each atom and butane, we're going to get a total molar mass of 58.12 g for one mole of butane. So we're going to be able to cancel out Graham's leaving us with moles. And this gives us a total of .344 moles of butane. And now, for our final step, we want to answer this question by finding our entropy of our reaction of our combustion reaction of butane. And so we begin with the heat that is absorbed by the water, the 991 kg jewels that we calculated. And we're going to divide that by our moles of our butane. So divided by .344 moles of butane. And so this gives us a value that is equal to 2880.8 kg joules per mole. So we want to recognize that the entropy of this combustion reaction is representative of the heat that this combustion reaction loses. Because this is technically an exo thermic reaction, it's losing heat, it's giving off heat. And so this is the heat that is lost in order for the water to absorb and become warmer. So we actually want to make this value negative for our final answer, since this reaction is XR thermic, and so to complete this example, this is going to be our final answer for our entropy of our reaction for the combustion of butane. So I hope that everything we reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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