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Thermodynamics question

2017 · Shift 0 · Q18
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Thermodynamics question

2017 · Shift 0 · Q18

JEE MainChemistryThermodynamicsMCQ+4 / −1
Given, C(graphite)+O2→CO2(g){C_{(graphite)}} + {O_2} \to C{O_2}(g)C(graphite)​+O2​→CO2​(g); ΔrHo{\Delta _r}{H^o}Δr​Ho= - 393.5 kJ mol-1 H2(g){{\rm H}_2}(g)H2​(g)+12O2(g)→H2O(l)ΔrHo{1 \over 2}{O_2}(g)\to {{\rm H}_2}{\rm O}(l){\Delta _r}{H^o}21​O2​(g)→H2​O(l)Δr​Ho= - 285.8 kJ mol-1 CO2(g)C{O_2}(g)CO2​(g)+2H2O(l)→CH4(g)2{{\rm H}_2}{\rm O}(l) \to C{H_4}(g)2H2​O(l)→CH4​(g)+2O2(g)ΔrHo2{O_2}(g){\Delta _r}{H^o}2O2​(g)Δr​Ho= + 890.3 kJ mol-1 Based on the above thermochemical equations, the value of ΔrHo{\Delta _r}{H^o}Δr​Ho at 298 K for the reaction C(graphite){C_{(graphite)}}C(graphite)​+2H2(g)→CH4(g)2{{\rm H}_2}(g) \to C{H_4}(g)2H2​(g)→CH4​(g) will be :
  1. A
    +144.0 kJ mol–1
  2. B
    – 74.8 kJ mol–1
  3. C
    -144.0 kJ mol–1
  4. D
    + 74.8 kJ mol–1
View written solutionFree

Correct answer: B

  1. Write the given thermochemical equations

We have:

(i)C(graphite)+O2→CO2(g)ΔH1∘=−393.5 kJ mol−1\text{(i)}\quad C_{(graphite)} + O_2 \to CO_2(g) \qquad \Delta H_1^\circ = -393.5\ \text{kJ mol}^{-1}(i)C(graphite)​+O2​→CO2​(g)ΔH1∘​=−393.5 kJ mol−1 (ii)H2(g)+12O2(g)→H2O(l)ΔH2∘=−285.8 kJ mol−1\text{(ii)}\quad H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l) \qquad \Delta H_2^\circ = -285.8\ \text{kJ mol}^{-1}(ii)H2​(g)+21​O2​(g)→H2​O(l)ΔH2∘​=−285.8 kJ mol−1 (iii)CO2(g)+2H2O(l)→CH4(g)+2O2(g)ΔH3∘=+890.3 kJ mol−1\text{(iii)}\quad CO_2(g) + 2H_2O(l) \to CH_4(g) + 2O_2(g) \qquad \Delta H_3^\circ = +890.3\ \text{kJ mol}^{-1}(iii)CO2​(g)+2H2​O(l)→CH4​(g)+2O2​(g)ΔH3∘​=+890.3 kJ mol−1

We need:

C(graphite)+2H2(g)→CH4(g)C_{(graphite)} + 2H_2(g) \to CH_4(g)C(graphite)​+2H2​(g)→CH4​(g)
  1. Use Hess's law

Add equations (i), twice equation (ii), and equation (iii).

First, multiply (ii) by 2:

2H2(g)+O2(g)→2H2O(l)ΔH∘=2(−285.8)=−571.6 kJ mol−12H_2(g) + O_2(g) \to 2H_2O(l) \qquad \Delta H^\circ = 2(-285.8) = -571.6\ \text{kJ mol}^{-1}2H2​(g)+O2​(g)→2H2​O(l)ΔH∘=2(−285.8)=−571.6 kJ mol−1

Now add:

C+O2→CO2C + O_2 \to CO_2C+O2​→CO2​ 2H2+O2→2H2O2H_2 + O_2 \to 2H_2O2H2​+O2​→2H2​O CO2+2H2O→CH4+2O2CO_2 + 2H_2O \to CH_4 + 2O_2CO2​+2H2​O→CH4​+2O2​
  1. Cancel common species

On adding, CO2CO_2CO2​ cancels with CO2CO_2CO2​, and 2H2O2H_2O2H2​O cancels with 2H2O2H_2O2H2​O. Also, total oxygen on left is O2+O2=2O2O_2 + O_2 = 2O_2O2​+O2​=2O2​, which cancels with 2O22O_22O2​ on the right.

So net reaction becomes:

C(graphite)+2H2(g)→CH4(g)C_{(graphite)} + 2H_2(g) \to CH_4(g)C(graphite)​+2H2​(g)→CH4​(g)

which is exactly the required reaction.

  1. Add enthalpy changes
ΔrH∘=(−393.5)+(−571.6)+890.3\Delta_r H^\circ = (-393.5) + (-571.6) + 890.3Δr​H∘=(−393.5)+(−571.6)+890.3 ΔrH∘=−965.1+890.3=−74.8 kJ mol−1\Delta_r H^\circ = -965.1 + 890.3 = -74.8\ \text{kJ mol}^{-1}Δr​H∘=−965.1+890.3=−74.8 kJ mol−1
  1. Match with options
ΔrH∘=−74.8 kJ mol−1\boxed{\Delta_r H^\circ = -74.8\ \text{kJ mol}^{-1}}Δr​H∘=−74.8 kJ mol−1​

So the correct option is B.

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