For which of the following reactions are ΔE and ΔH equal?
(a) CO2(g) + H2O(l) → H2CO
(b) 2 NaHCO3 (s) → Na2CO3(s) + H2O(g) + CO2(g)
(c) 2 H2(g) + O2(g) → 2 H2O(g)
(d) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Question

For which of the following reactions are ΔE and ΔH equal? 
(a) CO2(g) + H2O(l) → H2CO
(b) 2 NaHCO3 (s) → Na2CO3(s) + H2O(g) + CO2(g)
(c) 2 H2(g) + O2(g) → 2 H2O(g)
(d) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Accepted Answer

hi everyone for this problem. It reads which of the following will have a greater entropy change than internal energy change. And were given four options. So the equation that we're going to want to reference to solve this problem is entropy change is equal to internal energy change plus pressure times change in volume. Okay. And with this we're going to reference the ideal gas law and the ideal gas law is P. V equals N R T. So when we rewrite this, it's going to be pressure times change in volume is equal to the change in moles, times R times T. Okay. And we're gonna pay attention to this change in moles here and what that means? This change in moles is the change in moles of the gas. Okay, so it's going to equal the moles of the gas products, minus the moles of the gas react ints. So when we rewrite this first equation here, we're going to replace our pressure times change in volume. We're going to replace that with our change in moles times R, times T. Okay, So what that's going to look like is our change in entropy is going to equal change in internal energy plus change and moles of the gas times are times T. So what we're going to do for each of the options given here is we're going to calculate the change in moles of the gas and the number that we get is going to tell us whether or not the internal energy change is going to be greater than or less than or equal to the internal energy change. So let's go ahead and calculate the change in moles for the gas for each of the answer choices given All right. So for answer choice number one, our change and moles of the gas is going to equal. So for our products we're going to take the moles of gas products minus the moles of gas reactant. So for our products, let's take a look at the first one. We have two moles of product minus for our reactant. Sui have two moles and one mole. So we have three moles total of reactant. So this gives us negative one. Okay, so what this means then is our entropy change is less than our internal energy change because this number is less than zero. Alright, So for the second one, our change in moles for the gas. Okay, so for our products, let's count how many moles of product we have. So we have of gasses. Remember we're looking at gasses. So we're going to go ahead and ignore the liquid And we're going to ignore this acquis so we have one mole of gas products and for our reactant we have an Aquarius and we have a solid. Okay, so no gas is so this is going to be zero. So we have positive one. Alright, so what this means then is our entropy change is greater than our internal energy change because this number is greater than zero. Alright, let's go ahead and move on to the next one for our third One we have we're looking for moles of gas products and moles of gas reactant. So our change in moles for the gasses. So we have one mole of this gas and one mole of this gas. So we have two moles total for products and for our reactant we have two moles total. So this is -2. So we get zero For our change in moles for the 3rd 1. So what this means then is our entropy change is equal to our internal energy change because this number is zero. Alright, and For our last one we'll go ahead and calculate our change in moles. So our changing moles for the gas gas is So let's take a look at the last one for our products we have and Aquarius. So we'll go ahead and ignore that. So we have one mole of gas products and for our reactant both of them are a quiz. So we have zero moles of gas react ints. So this comes out to positive one. Okay, and so what this means is our change. Our entropy change is going to be greater than our internal energy change because this number is greater than zero. So the question asks us which of the following will have a greater entropy change than internal energy change. So we're looking for our positive values. So we see here we have a positive value for statement two and a positive value for statement for greater than zero value. So when we go back up to our answer choices, we can go ahead and erase all of this are correct. Answers are going to be Number two and # four. Okay, so let's go ahead and highlight that. So number two and number four have a greater entropy change than internal energy change. That is it for this problem? I hope this was helpful.

For which of the following reactions are ΔE and ΔH equal? (a) CO2(g) + H2O(l) → H2CO (b) 2 NaHCO3 (s) → Na2CO3(s) + H2O(g) + CO2(g) (c) 2 H2(g) + O2(g) → 2 H2O(g) (d) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Verified Solution

Key Concepts

Internal Energy (ΔE)

Enthalpy (ΔH)

Conditions for ΔE = ΔH