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Problem

Answer published by the source. Consult the official source to check your work against its answer.

question

For the reaction C(\mathrm{C}( graphite $)+\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons$ $\mathrm{CO}(g)+\mathrm{H}_2(g), \Delta H_R^{\circ}=131.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298.15 \mathrm{~K}$. Use the values of CP,m∘C_{P, m}^{\circ} at $298.15 \mathrm{~K}$ in the data tables to calculate ΔHR∘\Delta H_R^{\circ} at 125.0∘C125.0^{\circ} \mathrm{C}.
Plain-text mathematical notation (without MathML)
For the reaction C( graphite $)+\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons$ $\mathrm{CO}(g)+\mathrm{H}_2(g), \Delta H_R^{\circ}=131.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298.15 \mathrm{~K}$. Use the values of C_(P,m)^(∘) at $298.15 \mathrm{~K}$ in the data tables to calculate ΔH_(R)^(∘) at 125.0^(∘)C.
Original LaTeX notation
For the reaction $\mathrm{C}($ graphite $)+\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons$ $\mathrm{CO}(g)+\mathrm{H}_2(g), \Delta H_R^{\circ}=131.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298.15 \mathrm{~K}$. Use the values of $C_{P, m}^{\circ}$ at $298.15 \mathrm{~K}$ in the data tables to calculate $\Delta H_R^{\circ}$ at $125.0^{\circ} \mathrm{C}$.

category

physics

options

  1. 126.3 kJ mol^{-1}
  2. 128.5 kJ mol^{-1}
  3. 135.0 kJ mol^{-1}
  4. 132.9 $\mathrm{~kJ} \mathrm{~mol}^{-1}$
  5. 137.2 kJ mol^{-1}
  6. 133.5 kJ mol^{-1}
  7. 130.0 kJ mol^{-1}
  8. 131.5 $\mathrm{~kJ} \mathrm{~mol}^{-1}$
  9. 129.7 $\mathrm{~kJ} \mathrm{~mol}^{-1}$
  10. 134.5 $\mathrm{~kJ} \mathrm{~mol}^{-1}$

src

scibench-thermo

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Source and history

upstream_id / Original problem ID
9776
task_id
7181842c-527d-53b6-8e95-c0bbb136559f
task_key
default--test--9776
task_revision_id
1
config / split
default / test

Official source

initial import