For the reaction, PCl3 + Cl2= PCl5, delta H = -x kJ. If the delta Hf°of PCl3 is -y kJ, what is delta Hf° PCl5?Ans:-(x+y) kJ

Question

Ayush gupta · Answer

Since delta H of formation of PCl3 is(-y KJ)Than, for diffusion is (y KJ) now it`s given that delta H for given reaction is (xKJ) Now net enthalpy of formation of PCl5 will be = -x-y

Om Parkash · Answer

delta H for overall reaction is -X = enthapy of formation of (products- reactants) -X= PCL5 – ( -Y) -(X+Y)= ENTHAPY OF FORMATION OF PCL5

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For the reaction, PCl3 + Cl2= PCl5, delta H = -x kJ. If the delta Hf°of PCl3 is -y kJ, what is delta Hf° PCl5?Ans:-(x+y) kJ

For the reaction, PCl3 + Cl2= PCl5, delta H = -x kJ. If the delta Hf°of PCl3 is -y kJ, what is delta Hf° PCl5?Ans:-(x+y) kJ

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For the reaction, PCl3 + Cl2= PCl5, delta H = -x kJ. If the delta Hf°of PCl3 is -y kJ, what is delta Hf° PCl5?Ans:-(x+y) kJ

For the reaction, PCl3 + Cl2= PCl5, delta H = -x kJ. If the delta Hf°of PCl3 is -y kJ, what is delta Hf° PCl5?Ans:-(x+y) kJ

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