When 12.5 g of NH4NO3 is dissolved in 150.0 g of water of 25.0 °C in a coffee cup calorimeter, the final temperature of the solution of 19.7 °C. Assume that the specific heat of the solution is the same as that of water, 4.18 J/(g•°C). What is the ΔH per mol of NH4NO3? (LO 9.10) NH4NO3 (s) → NH4+ (aq) + NO3−(aq) ΔH = ? (a) +3.60 kJ (b) +23.0 kJ (c) +21.3 kJ (d) −3.60 kJ

Video transcript

All right. Hi, everyone. So this question says that when 12.5 g of ammonium nitrate is dissolved in 150.0 g of water of 25.0 °C in a coffee cup calorimeter. The final temperature of the solution is 19.7 °C. Assume that the specific heat of the solution is the same as that of water. 4.18 joules per gram degree Celsius. What is the delta H per mole of ammonium nitrate? And so here, our balanced chemical equation states that one mole of ammonium nitrate as a solid dissociates into one mole of ammonium ion in aqueous solution and one mole of nitrate ion in aqueous solution. And here we have four different answer choices for different values of delta H per mole. So first recall first and foremost that the delta H can be calculated by finding the heat of the reaction and dividing that by end, which is the number of moles. So first, let's find the heat of the reaction. Now recall that there is a formula for the heat or in other words, cue of a solution in queue of solution is equal to MC delta T where M is the mass C is the heat capacity or sorry specific heat. And delta T is the change in temperature of the process. Now, it just so happens that cue of the reaction is equal to negative Q solution, which means that the Q reaction is equal to negative MC delta T. So this is the formula that, that we're going to use for the first part of the equation, right. So Q of the reaction is equal to negative one multiplied by the mass of the system. So here in terms of the mass, we have 12.5 g of ammonium nitrate added to 150.0 g of water. So the sum of those compounds would be my total mass. Move this off to the side. Actually c refers to the specific heat which we assume to be 4.18 joules per gram degree Celsius and delta T is the difference between the final and initial temperatures. So the final temperature was 19.7 °C subtracted by the initial temperature which was 25.0 °C. So after evaluating this expression cue of the reaction or the heat of the reaction is equal to 3600 0.025 jules. All right. So now that we have the heat of the reaction, we're going to now find the number of moles of ammonium nitrate because recall that delta H would be equal to Q of the reaction divided by the number of moles. So in terms of n right, recall that the number of moles would be equal to the mass of ammonium nitrate divided by the molar mass of ammonium nitrate. So in this case, we have 12.5 g of ammonium nitrate and we will divide this by the molar mass. Now, the molar mass of ammonium nitrate is 80.043 g per mole. And so 12.5 g divided by 80.043 g per mole equals 0.1561 66 miles of ammonium nitrate. Now, I'm keeping the ungrounded value at least for right now to make my next calculation as precise as possible. So at this point, our final step is the calculation of delta H. So delta H is equal to cue of the reaction. That's 3600 0.025 jules divided by the number of moles which was 0.156166. And so my quotient is equal to 23,052 0.544 jewels per mole of ammonium nitrate. However, because all of our units are expressed in units of kilojoules. Excuse me, all of our answer choices are expressed in units of kilojoules. In this case, we're going to go ahead and divide our answer by 1000 or move the decimal place three spaces to the left. And so our final answer would be 23.0 kg joules per mole of ammonium nitrate, rounded to three significant figures and there you have it. So our answer is positive 23.0 kilojoules which corresponds with option B in the multiple choice. So if you stuck around until the end of this video, thank you so very much for watching and I hope you found this helpful.