Consider the following heat of formation data: H2O(l) … (2023)

Consider the following heat of formation data: H2O(l) Hfo =-285.83 kJ/mol C6H12O6(s) Hfo = -1273.06 kJ/mol Calculate Horxn forthe reaction below. Pick the choice which gives this Ho value.C6H12O6(s) 6 C(graphite) + 6 H2O(l) a) Horxn = -441.9 kJ b) Horxn =441.9 kJ c) Horxn = 1558.9 kJ d) Horxn = 987.2 kJ e) Horxn = -987.2kJm

Consider the following heat of formation data: H2O(l) Hfo = -285.83 kJ/mol C6H12O6(s) Hfo = -1273.06 kJ/mol Calculate Horxn for the reaction below. Pick the choice which gives this Ho value. C6H12O6(s) 6 C(graphite) + 6 H2O(l) a) Horxn = -441.9 kJ b) Horxn = 441.9 kJ c) Horxn = 1558.9 kJ d) Horxn = 987.2 kJ e) Horxn = -987.2 kJm

Consider the following heat of formation data: H2O(l) … (1)

Chemistry 101

(Video) Hess's Law Problems & Enthalpy Change - Chemistry

Determine the enthalpy of formation of $\mathrm{B}_{2} \mathrm{H}_{6}(g)$ in $\mathrm{kJ} / \mathrm{mol}$ of the following reaction. $$ \mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ Given : $\Delta_{r} H^{\circ}=-1941 \cdot \mathrm{kJ} / \mathrm{mol} ; \quad \Delta H_{f}^{\circ}\left(\mathrm{B}_{2} \mathrm{O}_{3}, s\right)=-1273 \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{W}_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}, g\right)=-241.8 \mathrm{~kJ} / \mathrm{mol}$ (a) $-75.6$ (b) $+7 \overline{5} .6$ (c) $-57.4$ (d) $-28.4$

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Hello students in this question we have to calculate the standard and healthy for the reaction. That is delta edge, not for the following reaction. Using the listed and copy of the reactions. That is delta H one not is equal to minus 4 75.8 kg jewel and delta H two not is equal to minus 3 55.0 kg. Do so we can write that the standards and therapy for the reaction will be equals two for the three by two. Delta H one not plus Delta H. Not to not and this one x 2. So we can substitute the values for three by two, multiplied by minus of 4 75.8 plus one by two times of minus 3 55.0. Okay, so after solving this value so Delta astronaut will be obtained as minus 713.7 minus 1 77.5. And after solving further this value is minus of 8 91.2. Hello. So from the given options option A. Here is the correct answer for this problem. Okay thank you.

Given that a plus B produced to see has Delta H equal to 600 killed jewels per mole and that to see plus DE to produce two e has adult each of 210 kill a Jules per mole. Let's determine the entropy change for each of the following for a four e Yeah, Thio foresee plus two d So we take to see Plus de put this to e Yeah, and we double that Delta each year is equal to two times 210 kill a Jules per mole, so this would be equal to four c plus two d is gonna be equal four e So Delta H here is equal to 420. Kill a Jules per mole and we're gonna flip this around. So four e produces foresee, and two D and Delta H years equals a negative 420 kill jewels per mole. So there is our answer for part a for part B a plus B plus de will produce two e and we need the entropy for this. So we know that a plus B to produce to see entropy here's equal to 600 killed jewels per mole and then to see plus de will produce two e. Delta H here is equal to 210 killer jewels per mole To see what council we would get a plus B plus D Do you produce to E and the Delta H share works out. Thio 810 Kill a Jules per mole. There's an answer for part B for C, we have see to produce one half a plus, one half be so if I take E plus B to produce to see and a half of this So the Delta H Year is equal to a half of that's 600 kill a Jules per mole. This is equal to half a plus a half B. Going to see Delta H here is equal to 300 killer jewels per mole. If we flip this around, see producer half a plus a half B. Delta H is equal to negative 300 kill a Jules per mole and there's our answer For part C for D, we have to see plus two e to produce to a plus to be plus de Okay, so this would yield right a plus B to produce to see we're gonna double this. So Delta H here is equal to two times 600 killer jewels per mole. So to a plus, to be to foresee Delta H years equals 1200 killer jewels promote and then we'll flip it around. Four c produced two A and two b Delta H year and sequel. The negative 1200 killer jewels per mole. And then we'll bring in to see Plus, de it would be equal Thio to e and we're gonna subtract. Well, Delta H here is equal to 2 10 killer jewels per mole and we're going to subtract thes so this would give us two C minus. D would produce to a plus to be minus two e delta H here would be equal to minus 14 10 killer jewels per mole. And then, if we rearrange here, this would be a to C plus two e would produce to a plus two B plus d. Delta H here is equal to minus 14 10 kill jewels. Permal. And here's our answer for part de and for E, we're going to determine the entropy change. E will produce a half a plus a half B plus a half de so starting from a plus B to produce to see Delta, each there to call the 600 killer jewels per mole and then to see plus DE to produce two e. Delta H here is equal to 10. Kill a Jules per mole. It's combined these together two seats for council we get a plus B plus D would be to e Delta eight shares equal to 810 killed jewels per mole. And if we take this and take half of this, we would get a half a plus a half B plus a half D on that would produce E. I thought the H here would be 405 killer jewels per mole. And then, if we flip this around, E would be producing half a plus a half B plus a half E and Delta H here would be negative 405 killer jewels per mole. There would be your answer for Part E. Yeah,

Hello students in this question we have given for the reaction Delta H is -3-7 1 kill a jew. So we have to determine that delta you for the conversion of 1.5 mold of Benji. Okay, So first of all, 41 mole of the bending, that number of moles change. It is equal to the the number of most in the uh reactant. That is one plus 15 by two and minus number of moles in the product that is six plus three and this is obtained as minus three by two or minus 1.5. So delta H 41 mole. This can be written as delta U plus delta N. G. Popular. RT so delta H is minus 3 to 71 kill a jew. This is equal to delta U plus N. G. Which is minus 1.5 Manipulated by our which is 8.314 and temperature T which is 27° integrated that is 300 Calvin and they were by thousands to convert this into a jew. So delta you. For the even more it is equal to -326-7.25. So for the 1.5 mold, This value will be equal to 1.5 times of delta you. Okay, so 1.5 player b minus of 3267.25. So delta U dash 41 bowl, it is equal to minus 49 double zero point double eight kg. Okay, so from the given options, option B is the correct answer for this problem. Thank you.

Hello students. In this question we have to calculate the entropy of formation for the H. 202 in the liquid form. Okay so we can write from the given accusations that is delta H. Formation For the H. 202. This will be equals to the delta edge for the third reaction plus one by two of delta reaction of H two formation minus one by two of delta H formation for the reaction of first reaction. Okay so we can substitute the values so we will get that minus 2 85 plus one by two. Medical herb minus six double too -1 by tumour player B -818. Okay, so after solving the delta X formation for the H. 202 in the liquid form is obtained as minus 1 87 kg jewel per boat. So it means from the given options option B. Here is the correct answer for the problem. Okay thank you.

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