consider the following intermediate chemical equations.\nch₄(g)→c(s)+2h₂(g) δh₁ = 74.6kj\nccl₄(g)→c(s)+2cl₂(g…

consider the following intermediate chemical equations.\nch₄(g)→c(s)+2h₂(g) δh₁ = 74.6kj\nccl₄(g)→c(s)+2cl₂(g) δh₂ = 95.7kj\nh₂(g)+cl₂(g)→2hcl(g) δh₃ = -92.3kj\nwhat is the enthalpy of the overall chemical reaction ch₄(g)+4cl₂(g)→ccl₄(g)+4hcl(g)?\n-205.7 kj\n-113.4 kj\n-14.3 kj\n78.0 kj

consider the following intermediate chemical equations.\nch₄(g)→c(s)+2h₂(g) δh₁ = 74.6kj\nccl₄(g)→c(s)+2cl₂(g) δh₂ = 95.7kj\nh₂(g)+cl₂(g)→2hcl(g) δh₃ = -92.3kj\nwhat is the enthalpy of the overall chemical reaction ch₄(g)+4cl₂(g)→ccl₄(g)+4hcl(g)?\n-205.7 kj\n-113.4 kj\n-14.3 kj\n78.0 kj

Answer

Explanation:

Step1: Reverse the second - equation

Reverse $CCl_4(g)\to C(s)+2Cl_2(g)$ to $C(s)+2Cl_2(g)\to CCl_4(g)$ and change the sign of $\Delta H_2$. So the new $\Delta H_2=- 95.7$ kJ.

Step2: Multiply the third - equation by 2

Multiply $H_2(g)+Cl_2(g)\to2HCl(g)$ by 2 to get $2H_2(g)+2Cl_2(g)\to4HCl(g)$ and $\Delta H_3' = 2\times(-92.3)\text{ kJ}=-184.6$ kJ.

Step3: Add the equations

Add $CH_4(g)\to C(s)+2H_2(g)$ ($\Delta H_1 = 74.6$ kJ), $C(s)+2Cl_2(g)\to CCl_4(g)$ ($\Delta H_2=-95.7$ kJ) and $2H_2(g)+2Cl_2(g)\to4HCl(g)$ ($\Delta H_3'=-184.6$ kJ) together. The sum of the left - hand sides is $CH_4(g)+4Cl_2(g)$ and the sum of the right - hand sides is $CCl_4(g)+4HCl(g)$. The sum of the enthalpies is $\Delta H=\Delta H_1+\Delta H_2+\Delta H_3'=74.6+( - 95.7)+(-184.6)$ kJ. $\Delta H=74.6 - 95.7-184.6=-205.7$ kJ.

Answer:

-205.7 kJ