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Use the Standard Reaction Enthalpies Given Below to Determine

question 78

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Use the standard reaction enthalpies given below to determine Use the standard reaction enthalpies given below to determine   H° for the following reaction: P<sub>4</sub>(g)  + 10Cl<sub>2</sub>(g)  → 4PCl<sub>5</sub>(s)    H°<sub> </sub>= ? Given: PCl<sub>5</sub>(s)  → PCl<sub>3</sub>(g)  + Cl<sub>2</sub>(g)    H°= +157 kJ P<sub>4</sub>(g)  + 6Cl<sub>2</sub>(g)  → 4PCl<sub>3</sub>(g)    H° = -1207 kJ A)  -1835 kJ mol<sup>-1</sup> B)  -1364 kJ mol<sup>-1</sup> C)  -1050 kJ mol<sup>-1</sup> D)  -1786 kJ mol<sup>-1</sup> E)  -2100 kJ mol<sup>-1</sup> H° for the following reaction: P4(g) + 10Cl2(g) → 4PCl5(s) Use the standard reaction enthalpies given below to determine   H° for the following reaction: P<sub>4</sub>(g)  + 10Cl<sub>2</sub>(g)  → 4PCl<sub>5</sub>(s)    H°<sub> </sub>= ? Given: PCl<sub>5</sub>(s)  → PCl<sub>3</sub>(g)  + Cl<sub>2</sub>(g)    H°= +157 kJ P<sub>4</sub>(g)  + 6Cl<sub>2</sub>(g)  → 4PCl<sub>3</sub>(g)    H° = -1207 kJ A)  -1835 kJ mol<sup>-1</sup> B)  -1364 kJ mol<sup>-1</sup> C)  -1050 kJ mol<sup>-1</sup> D)  -1786 kJ mol<sup>-1</sup> E)  -2100 kJ mol<sup>-1</sup> = ?
Given:
PCl5(s) → PCl3(g) + Cl2(g) Use the standard reaction enthalpies given below to determine   H° for the following reaction: P<sub>4</sub>(g)  + 10Cl<sub>2</sub>(g)  → 4PCl<sub>5</sub>(s)    H°<sub> </sub>= ? Given: PCl<sub>5</sub>(s)  → PCl<sub>3</sub>(g)  + Cl<sub>2</sub>(g)    H°= +157 kJ P<sub>4</sub>(g)  + 6Cl<sub>2</sub>(g)  → 4PCl<sub>3</sub>(g)    H° = -1207 kJ A)  -1835 kJ mol<sup>-1</sup> B)  -1364 kJ mol<sup>-1</sup> C)  -1050 kJ mol<sup>-1</sup> D)  -1786 kJ mol<sup>-1</sup> E)  -2100 kJ mol<sup>-1</sup> H°= +157 kJ
P4(g) + 6Cl2(g) → 4PCl3(g) Use the standard reaction enthalpies given below to determine   H° for the following reaction: P<sub>4</sub>(g)  + 10Cl<sub>2</sub>(g)  → 4PCl<sub>5</sub>(s)    H°<sub> </sub>= ? Given: PCl<sub>5</sub>(s)  → PCl<sub>3</sub>(g)  + Cl<sub>2</sub>(g)    H°= +157 kJ P<sub>4</sub>(g)  + 6Cl<sub>2</sub>(g)  → 4PCl<sub>3</sub>(g)    H° = -1207 kJ A)  -1835 kJ mol<sup>-1</sup> B)  -1364 kJ mol<sup>-1</sup> C)  -1050 kJ mol<sup>-1</sup> D)  -1786 kJ mol<sup>-1</sup> E)  -2100 kJ mol<sup>-1</sup> H° = -1207 kJ

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