Gaokao Chemistry Practice Question

2.（3 分）下面均是正丁烷与氧气反应的热化学方程式 $\left(25^{\circ}, 101 \mathrm{kPa} ）:\right.$ (1) $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{CO}_{2}(\mathrm{~g})+5 \mathrm{H}_{2} \mathrm{O}$ (1) $\triangle \mathrm{H}=-2878 \mathrm{~kJ} / \mathrm{mol}$ (2) $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{CO}_{2}(\mathrm{~g})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \triangle \mathrm{H}=-2658 \mathrm{~kJ} / \mathrm{mol}$ (3) $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+\frac{9}{2} \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{CO}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}$ (1) $\triangle \mathrm{H}=-1746 \mathrm{~kJ} / \mathrm{mol}$ (4) $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+\frac{9}{2} \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{CO}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \triangle \mathrm{H}=-1526 \mathrm{~kJ} / \mathrm{mol}$ 由此判断，正丁烷的燃烧热是（ $）$

1. A.$-2878 \mathrm{~kJ} / \mathrm{mol}$
2. B.$-2658 \mathrm{~kJ} / \mathrm{mol}$
3. C.$-1746 \mathrm{~kJ} / \mathrm{mol}$
4. D.$-1526 \mathrm{~kJ} / \mathrm{mol}$