Calculate ΔHrxn for the following reaction: CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) given these reactions and their ΔH values: C(s)C(s)H2(g)+++2H2(g)2Cl2(g)Cl2(g) →→→CH4(g),CCl4(g),2HCl(g),ΔH=−74.6 kJΔH=−95.7 kJΔH=−184.6 kJ

∆Hrxn = -390.3 kJ.

Explanation:

Okay, let us first write out the additional equations given below;

C(s) + 2 H2(g) > CH4(g); ΔH= −74.6kJ (1).

C(s) + 2 Cl2(g) > CCl4(g); ΔH=−95.7kJ (2).

H2(g) + Cl2(g) > 2HCl(g); ΔH=−184.6kJ (3).

Step one: reverse equation (1).

This gives us equation (4) which is;

CH4 > C + 2H2 ; ∆H = + 74.6 kJ (4).

Step two: multiply equation (3) by 2 to get equation (5) which is;

2 H2(g) + 2 Cl2(g) > 4 HCl(g); ΔH=− 369.2 kJ. (5).

Step three: addition of equation (4), (2) and (5) gives us our desired equation;

CH4(g) + 4 Cl2(g) > CCl4(g) + 4 HCl(g).

Step four:

Addition of all the ∆H gives us our answer;

(+74.6)kJ + (-95.7)kJ + (− 369.2)kJ.

Therefore, ∆Hrxn = -390.3 kJ.

the answer is the

it would be d. henry moseley

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