Using the following reactions, calculate the ΔH,for the following
reaction, the preparation of the unstable acid nitrous acid. Please show full solutions!
Thank you, informations are provided below
HCl(g) + NaNO2(s) → HNO2(l) + NaCl(s) ΔH° = ?
i. 2NaCl(s) + H2O(l) → 2 HCl(g) + Na2O(s) ΔH° = +507.31 kJ
ii. NO(g) + NO2(g) + Na2O(s)→ 2NaNO2(s) ΔH° = -427.14 kJ
iii. NO(g) + NO2(g) → N2O(g) + O2(g) ΔH° = -42.68 kJ
iv. 2HNO2(l) → N2O(g) + O2(g) + H2O(l) ΔH° = + 34.35 kJ
Reverse equation 1, reverse equation 2, leave equation 3 alone and reverse equation 4. Then add all of them together. When you reverse an equation you change dH sign. Note that all of the extraneous components cancel. The final equation will be twice what you want; therefore, divide the dH you obtain by 2. Post ALL of your work if you get stuck. The equation below show 1, 2, 4 reversed but I did NOT show the sum.
<i. 2HCl(g) + Na2O(s) ==>2NaCl(s) + H2O(l)
<ii. 2NaNO2(s) ==> NO(g) + NO2(g) + Na2O(s)
>iii. NO(g) + NO2(g) → N2O(g) + O2(g)
<iv. N2O(g) + O2(g) + H2O(l) ==> 2HNO2(l)
HCl(g) + NaNO2(s) → HNO2(l) + NaCl(s) is the final equation you want.