Calculate Delta H in kJ for the following rea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the enthalpy change for the reaction $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$, we use Hess's Law by adding the given equations:$(i)

Vedclass · Accepted Answer

$-393$

Vedclass · Answer

(A) To find the enthalpy change for the reaction $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$, we use Hess's Law by adding the given equations:$(i) \ H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H_1 = +131 \ kJ$$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H_2 = -282 \ kJ$$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H_3 = -242 \ kJ$Adding equations $(i)$, $(ii)$, and $(iii)$:$(H_2O_{(g)} + C_{(s)} + CO_{(g)} + \frac{1}{2} O_{2(g)} + H_{2(g)} + \frac{1}{2} O_{2(g)})$ $\longrightarrow (CO_{(g)} + H_{2(g)} + CO_{2(g)} + H_2O_{(g)})$Canceling common species on both sides, we get:$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = 131 + (-282) + (-242) = -393 \ kJ$.

Similar Questions

State Hess's Law of Constant Heat Summation.

For the reaction $R \rightarrow P$,the following potential energy diagram is given. What will be the enthalpy change $(\Delta H)$ for the given reaction (in $kJ$)?

One mole of acetone requires less heat to vaporise than $1$ mol of water. Which of the two liquids has higher enthalpy of vaporisation?

What is the enthalpy of formation of $NH_3$ if the bond enthalpies are $(N \equiv N) = 941 \ kJ/mol$,$(H-H) = 436 \ kJ/mol$,and $(N-H) = 389 \ kJ/mol$?

The heats of formation of $CO_2$ and $H_2O$ are $-97 \, kcal$ and $-68 \, kcal$ respectively. The heat of combustion of benzene is $-783 \, kcal$. What is the heat of formation of benzene in $kcal$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module