Calculate the standard enthalpy of the reaction :SiO2(s) + 3C(graphite) → SiC(s) + 2CO(g) from the following reactions :(i) Si(s) + O2(g) → SiO2(s), ΔrH° = −911 kJ(ii) 2C(graphite) + O2(g) → 2CO(g), ΔrH° = −221 kJ(iii) Si(s) + C(graphite) → SiC(s), Δ…

Q.18 Calculate the standard enthalpy of the reaction :

SiO_2(s) + 3C_(graphite) → SiC_(s) + 2CO_(g) from the following reactions :

(i) Si_(s) + O_2(g) → SiO_2(s), Δ_rH° = −911 kJ

(ii) 2C_(graphite) + O_2(g) → 2CO_(g), Δ_rH° = −221 kJ

(iii) Si_(s) + C_(graphite) → SiC_(s), Δ_rH° = −65.3 kJ

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: \(\Delta_r H^\circ = +624.7\) kJ

The required equation is obtained by reversing (i) and adding (ii) and (iii):

Reverse (i): SiO_2(s) → Si_(s) + O_2(g), \(\Delta H^\circ = +911\) kJ

(ii): 2C_(graphite) + O_2(g) → 2CO_(g), \(\Delta H^\circ = -221\) kJ

(iii): Si_(s) + C_(graphite) → SiC_(s), \(\Delta H^\circ = -65.3\) kJ

Adding (Si and O₂ cancel): SiO_2(s) + 3C_(graphite) → SiC_(s) + 2CO_(g)

\(\Delta_r H^\circ = 911 + (-221) + (-65.3) = +624.7\) kJ

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

Was this answer helpful?