1.(2024·河北衡水高三檢測(cè))硫酸工業(yè)中的釩催化劑參與反應(yīng)的相對(duì)能量變化如圖所示,下列熱化學(xué)方程式錯(cuò)誤的是( B )
A.V2O4(s)+SO3(g)V2O5(s)+SO2(g) ΔH=-24 kJ·mol-1
B.2VOSO4(s)V2O4(s)+2SO3(g) ΔH=+200 kJ·mol-1
C.2V2O5(s)+2SO2(g)2VOSO4(s)+V2O4(s) ΔH=-352 kJ·mol-1
D.V2O5(s)+SO2(g)+SO3(g)2VOSO4(s) ΔH=-376 kJ·mol-1
[解析] 由反應(yīng)物總能量高于生成物總能量可知,為放熱反應(yīng),則V2O4(s)+SO3(g)V2O5(s)+SO2(g) ΔH=-24 kJ·mol-1,A項(xiàng)正確;途徑②的ΔH=+[24-(-176)]kJ·mol-1=+200 kJ·mol-1,2VOSO4(s)V2O4(s)+2SO3(g) ΔH=+400 kJ·mol-1,B項(xiàng)錯(cuò)誤;途徑①的反應(yīng)物總能量大于生成物總能量,為放熱反應(yīng),2 mol V2O5反應(yīng)放出能量為2×176=352 kJ,則2V2O5(s)+2SO2(g)2VOSO4(s)+V2O4(s) ΔH=-352 kJ·mol-1,C項(xiàng)正確;根據(jù)蓋斯定律,由反應(yīng)V2O4(s)+SO3(g)V2O5(s)+SO2(g) ΔH=-24 kJ·mol-1加上2VOSO4(s)V2O4(s)+2SO3(g) ΔH=+400 kJ·mol-1可得2VOSO4(s)V2O5(s)+SO2(g)+SO3(g) ΔH=+376 kJ·mol-1,則V2O5(s)+SO2(g)+SO3(g)2VOSO4(s) ΔH=-376 kJ·mol-1,D項(xiàng)正確。